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THE   MODERN   MATHEMATICAL   SERIES 

LUCIEN   AUGUSTUS   WAIT  .  .  .  General  Editor 

(SEKIOK  PBOFESSOB  OF  MATHEMATICS  IN  COBNELL  UNIVBBSITY) 


The  Modern  Mathematical  Series, 
lucien  augustus  wait, 

(Senior  Professor  of  Mathematics  in  Cornell  University,) 
GENERAL  EDITOR. 


This  series  includes  the  followiug  works : 

ANALYTIC  GEOMETRY.    By  J.  H.  Tanner  and  Joseph  Allen. 
DIFFERENTIAL  CALCULUS.    By  James  McMahon  and  Virgil  Snyder. 
INTEGRAL  CALCULUS.    By  D.  A.  Murray. 
DIFFERENTIAL  AND  INTEGRAL  CALCULUS.    By  Virgil  Snyder  and  J.  I. 

Hutchinson. 
ELEMENTARY  ALGEBRA.    By  J.  H.  Tanner. 
ELEMENTARY  GEOMETRY.    By  James  McMahon. 


The  Analytic  Geometry,  Differential  Calculus,  and  Integral  Calculus  (pub- 
lished in  September  of  1898)  were  written  primarily  to  meet  the  needs  of  college 
students  pursuing  courses  in  Engineering  and  Architecture ;  accordingly,  prac- 
tical problems,  in  illustration  of  general  principles  under  discussion,  play  an 
important  part  in  each  book. 

These  three  books,  treating  their  subjects  in  a  way  that  is  simple  and  practi- 
cal, yet  thoroughly  rigorous,  and  attractive  to  both  teacher  and  student,  received 
such  general  and  hearty  approval  of  teachers,  and  have  been  so  widely  adopted 
in  the  best  colleges  and  universities  of  the  country,  that  other  books,  written  on 
the  same  general  plan,  are  being  added  to  the  series. 

The  Differential  and  Integral  Calculus  in  one  volume  was  written  especially 
for  those  institutions  where  the  time  given  to  these  subjects  is  not  sufficient  to 
use  advantageously  the  two  separate  books. 

The  more  elementary  books  of  this  series  are  designed  to  implant  the  spirit  of 
the  other  books  into  the  secondary  schools.  This  will  make  the  work,  from  the 
schools  up  through  the  university,  continuous  and  harmonious,  and  free  from 
the  abrupt  transition  which  the  student  so  often  experiences  in  changing  from 
his  preparatory  to  his  college  mathematics. 


ELEMENTARY  GEOMETRY 


PLANE 


BY 


JAMES   McMAHON 

( I 

ASSISTANT  PROFESSOR  OF  MATHEMATICS  IN   CORNELL  UNIVERSITY 


>>©<€ 


NEW  YORK.:.  CINCINNATI.:.  CHICAGO 

AMERICAN    BOOK    COMPANY 


COPYEIGHT,   1903,   BY 

JAMES  McMAHON. 

McM.  Elem.  Geom. 
W.  p.  I 


PREFACE 


This  text-book  aims  to  carry  out  the  spirit  of  the  admi- 
rable suggestions  made  by  the  Committee  on  Secondary 
School  studies,  appointed  by  the  National  Educational  Asso- 
ciation. While  the  book  speaks  for  itself,  some  of  its  lead- 
ing features  may  here  be  pointed  out. 

(1)  It  aims  at  a  combination  of  Euclidean  rigor  with 
modern  methods  of  presentation  suitable  for  beginners  in 
the  study  of  demonstrative  geometry ;  but  the  rigor  is  not 
regarded  as  consisting  so  much  in  excessive  formality  of 
expression  as  in  soundness  of  structural  development. 

(2)  It  regards  the  postulates  as  a  body  of  fundamental 
conventions  that  constitute  a  definition  of  Euclidean  space, 
from  which  (with  the  definitions  of  particular  figures)  other 
properties  of  such  space  are  to  be  unfolded  by  a  series  of 
logical  steps. 

(3)  It  regards  the  postulates  of  construction  as  determin- 
ing or  defining  the  province  of  elementary  as  distinguished 
from  higher  geometry.  Accordingly  no  hypothetical  figure 
is  made  the  basis  of  an  argument  until  its  construction  has 
been  proved  to  be  reducible  to  the  construction  postulates; 
and  thus  problems,  no  less  than  theorems,  have  their  place 
in  the  logical  development  of  the  subject. 

(4)  The  theorems  and  problems  are  arranged  in  natural 
groups  and  subgroups  with  reference  to  their  underlying 
principle,  thus  exhibiting  the  gradual  unfolding  of  the 
space  relations. 

(5)  Elementary  ideas  of  logic  are  introduced  comparatively 
early,  so  that  the  student  may  easily  recognize  the  equiva- 


-Z€yr%£yr^rk^ 


vi  PREFACE 

lence  of  statements  that  differ  only  in  form,  and  also  dis- 
tinguish between  different  statements  that  may  seem  to  be 
alike. 

(6)  The  mode  of  treating  ordinary  size-relations  is  purely 
geometrical.  "  This  method  being  pure  and  thoroughly  ele- 
mentary, and  involving  no  abstraction,  is  surely  better  suited 
to  the  beginner.  Indeed,  the  student  is  most  likely  to  be- 
come a  sound  geometer  who  is  not  introduced  to  the  notion 
of  numerical  measures  until  he  has  learned  that  geometry 
can  be  developed  independently  of  it  altogether.  For  this 
notion  is  subtle,  and  highly  artificial  from  a  purely  geomet- 
rical point  of  view,  and  its  rigorous  treatment  is  difficult. 
The  student  generally  only  half  comprehends  it,  so  that  for 
him  demonstrations  lose  more  in  rigor  as  well  as  in  vivid- 
ness and  objectivity  by  its  use  than  they  gain  in  apparent 
simplicity.  Moreover,  the  constant  association  of  number 
with  the  geometric  magnitudes  as  one  of  their  properties, 
tends  to  obscure  the  fundamental  characteristic  of  these 
magnitudes  —  their  continuity."*  Words  suggestive  of 
measurement,  such  as  length,  area,  distance,  etc.,  are 
accordingly  not  used  in  the   purely  geometrical  chapters. 

(7)  The  Euclidean  doctrine  of  ratio  and  proportion  is  pre- 
sented in  a  modernized  form,  which  shows  its  naturalness 
and  generality,  and  renders  it  easier  of  application  than  the 
unsatisfactory  numerical  theory  which  is  so  often  allowed 
to  usurp  its  place,  although  it  is  generally  conceded  by 
mathematicians  that  Euclid's  treatment  of  proportion  is 
one  of  the  most  admirable  and  beautiful  of  his  contribu- 
tions to  geometry. 

(8)  There  is  a  chapter  on  mensuration,  in  which  measure- 
numbers  are  introduced  as  a  natural  outgrowth  from  the 
general  notion  of  ratio,  and  the  irrational  numbers  that  cor- 

*  See  Report  of  Conference  of  School  and  College  Teachers  em- 
bodied in  the  Report  of  the  Committee  of  Ten,  p.  113.  (Published 
for  the  National  Educational  Association  by  American  Book  Company, 
1894.) 


PREFACE  vii 

respond  to  the  ratios  of  incommensurable  magnitudes  are 
given  simple  logical  treatment  based  on  the  general  theory 
of  ratio,  without  resorting  to  the  notion  of  a  limit,  which 
has  no  natural  connection  with  the  subject. 

(9)  The  measurement  of  the  circle  is  based  on  the  cor- 
rect definition  of  the  length  of  a  curved  line  (in  terms 
of  a  straight  measuring-unit)  given  by  the  best  continental 
writers.  Here  the  idea  of  a  limit  is  imbedded  in  the  defi- 
nition ;  but  the  existence  and  uniqueness  of  the  limit  must 
be  proved  before  we  can  speak  of  the  "  length  of  an  arc  "  so 
as  to  make  it  the  subject  of  our  discourse ;  otherwise  we  are 
using  a  word  that  has  not  been  completely  defined.  Similar 
statements  may  be  made  with  regard  to  the  area  of  the  circle. 
As  far  as  the  author  is  aware  this  plan  has  not  hitherto 
been  followed  in  any  text-book  in  the  English  language.  It 
is  hoped  that  this  important  topic  has  been  presented  in  a 
rigorous  and  simple  manner. 

(10)  Throughout  the  book  there  is  an  endeavor  to  develop 
the  student's  power  of  invention  and  generalization,  with- 
out encouraging  looseness,  or  introducing  discouraging  diffi- 
culties. 

These  features  have  received  the  approval  of  several  ex- 
perienced educators.  Special  acknowledgments  are  due  to 
Professors  Wait,  Jones,  Tanner,  and  Stecker  for  assistance 
and  advice. 


SUGGESTIONS   TO   TEACHERS 

It  is  suggested  to  teachers  that  the  introductory  articles 
be  read  and  discussed  in  class  in  an  informal  way,  with  the 
aim  of  drawing  out  and  clarifying  those  ideas  of  space- 
relations  which  the  students  may  already  possess.  Some 
of  the  introductory  matter  can  be  passed  over  lightly  on 
first  reading,  and  returned  to  when  necessary.  Teachers 
may  exercise  their  discretion  with  regard  to  articles  in  small 
print  throughout  the  book. 

For  a  shorter  course,  any  of  the  following  groups  of  arti- 
cles may  be  omitted  without  breaking  the  continuity  of  the 

subject :  — 

Book  I.  180-186,  195-213,  232-247. 

Book  II.  2-3,  79-88,  90-107. 

Book  III.  141-198. 

Book  IV.  10. 

Most  of  the  exercises  that  are  given  in  immediate  connec- 
tion with  the  propositions  should  be  solved  by  the  student; 
but  only  a  few  of  those  placed  at  the  end  of  sections  need 
be  taken  on  a  first  reading.  They  are  all  carefully  graded, 
and  many  suggestions  are  given.  The  author  will  be  glad 
to  hear  from  any  person  who  may  meet  with  any  error  or 
difficulty. 

As  some  teachers  may  wish  to  use  the  Socratic  or  heuristic 
methods  of  instruction  in  certain  parts  of  the  work,  the 
arrangement  and  development  of  the  topics  are  such  as 
to  lend  themselves  easily  to  these  valuable  pedagogical 
methods,  without  interfering  with  the  more  formal  presen- 
tation that  is  appropriate  to  a  course  in  demonstrative 
geometry.  The  actual  details  of  any  such  method  are,  how- 
ever, left  to  individual  discretion,  as  the  skillful  teacher  has 
usually  no  difficulty  in  reconciling  the  claims  of  pedagogy 
and  sound  reasoning. 

viii 


CONTENTS 
INTRODUCTION 

PAGE 

The  Fundamental  Space  Concepts 1 

Primary  Space  Postulates 2 

Primary  Definitions 3 

The  Construction  Postulates 7 

Primary  Magnitude  Relations 9 

PLANE   GEOMETEY 

BOOK   I.  —  Rectilinear  Figures 

Line-segments  and  Angles 10 

Axioms  concerning  Lines  and  Angles 16 

Some  Logical  Terms 18 

Theorems  concerning  Angles 20 

Triangles 25 

Some  Fundamental  Constructions 29 

Summary  of  Types  of  Inference 45 

Parallel  Lines 53 

Construction  of  Triangles 62 

Quadrangles 72 

Polygons 89 

Symmetry     .  - 104 

Locus  Problems 112 

Methods  of  Analysis 120 

BOOK  XL — Equivalence  of  Polygons 

General  Principles 125 

Comparison  of  Parallelograms 130 

Equivalences  involving  Rectangles        .         .        .        .        .        .137 

Equivalences  in  a  Triangle 147 

Construction  of  Equivalent  Polygons 154 

Division  of  a  Line 1^9 

Locus  Problems ^^^ 

Maxima  and  Minima 1^^ 


X  CONTENTS 

BOOK  III.  —  The  Circle 

PAGE 

Fundamental  Properties 1^^ 

Properties  of  Equal  Circles 174 

Angles  in  Segments 186 

Tangents 1^2 

Two  Circles 1^^ 

Concurrent  Chords 206 

Inscription  and  Circumscription 208  . 

Maxima  and  Minima 222 

Locus  Problems 235 

BOOK  IV.  — Eatio  and  Proportion 

Multiples  and  Measures 242 

Scale  of  Relation 251 

On  the  Notion  of  Ratio 252 

Properties  of  Ratios 255 

Properties  of  a  Proportion 264 

Two  or  More  Proportions 267 

BOOK  V.  —  Ratios  of  Lines,  Polygons,  etc. 

Similarly  Divided  Lines 271 

Compounding  of  Ratios 279 

Similar  Triangles 285 

Similar  Polygons 295 

Surface  Ratios 304 

Ratios  in  the  Circle 315 

Locus  Problems ,     .        .  322 

BOOK  VI.  —  Mensuration 

Abbreviated  Scale 325 

Associated  Numerical  Ratios 328 

Number-correspopdent 330 

Irrational  Numbers 332 

Measure-number 334 

Measurement  of  Rectangles .        . 337 

Directed  Lines 339 

Measurement  of  Triangles 341 

Measurement  of  Regular  Polygons 344 

Measurement  of  the  Circle  —  Variables  and  Limits      .        .        .  346 


ELEMENTARY  GEOMETRY 

INTRODUCTION 

The  Four  Fundamental  Space  Concepts 

1.  Geometry  is  that  branch  of  mathematical  science  which 
treats  of  the  properties  of  space. 

The  space  in  which  we  live  is  divisible  into  parts.  Every 
portion  of  matter  occupies  a  part  of  space.  The  portion  of 
space  occupied  by  a  body,  considered  separately  from  the 
matter  which  it  contains,  may  be  regarded  as  existing 
unchanged  when  the  body  moves  into  another  portion  of 
space. 

2.  Any  portion  of  space  capable  of  being  occupied  by  a 
physical  solid  is  called  a  geometrical  solid,  or  simply  a 
solid. 

3.  The  common  boundary  of  two  adjoining  solids,  or  of  a 
solid  and  the  surrounding  space,  is  not  a  solid  ;  it  is  a  second 
kind  of  space  element,  called  a  surface. 

4.  Any  surface  is  likewise  divisible  into  parts;  and  the 
common  boundary  of  two  adjoining  parts  of  a  surface  is  not 
a  surface ;  it  is  a  third  kind  of  space  element,  called  a  line. 

5.  Again,  any  line  is  divisible  into  parts  ;  and  the  common 
extremity  of  two  adjoining  parts  of  a  line  is  a  fourth  kind 
of  space  element,  called  a  point. 

A  point  is  not  divisible  into  parts ;  hence,  the  point  is  the 
simplest  space  element. 

6.  A  fine  tracing  point,  or  a  dot  on  a  sheet  of  paper,  gives 
an  approximate  representation  of  the  ideal  geometric  point. 


2  INTRODUCTION 

Similarly  the  lines  which  we  trace  on  the  surface  of  a 
sheet  of  paper  give  some  idea  of  geometric  lines.  They  are, 
however,  only  approximations  to  ideal  geometric  lines,  no 
matter  how  finely  they  may  be  traced. 

7.  It  is  often  convenient  to  think  of  a  geometric  line  as 
traced  or  generated  by  a  point  of  a  moving  body. 

The  line  is  then  called  the  path  of  the  point. 

In  the  same  way  a  surface  may  be  imagined  as  generated 
by  a  line  that  is  traced  on  a  moving  body. 

Again,  the  surface  of  a  moving  body  may  be  imagined  as 
tracing  or  sweeping  out  a  solid  portion  of  space. 

We  cannot  go  on,  however,  and  imagine  any  motion  of  a 
solid  that  will  generate  any  higher  space  concept. 

Hence,  the  solid  is  the  most  comprehensive  space  concept 
we  can  form. 

8.  Thus  whether  we  begin  with  the  notion  of  a  solid  and 
proceed  downwards  to  the  notion  of  a  point,  or  whether  we 
begin  with  the  point  and  build  up  the  solid,  there  are  but 
three  steps  in  the  process :  from  solid  to  surface,  surface  to 
line,  line  to  point ;  or  else  from  point  to  line,  line  to  surface, 
surface  to  solid. 

Accordingly,  the  space  of  our  experience  is  said  to  have 
three  dimensions. 

A  point  is  said  to  have  no  extension  and  no  dimensions;  a 
line  is  said  to  be  extended  in  one  dimension ;  a  surface  to  be 
extended  in  two  dimensions ;  a  solid  to  be  extended  in  three 
dimensions. 

9.  Any  combination  of  points,  lines,  surfaces,  or  solids,  is 
called  a  geometric  figure. 

Two  Primary  Space  Postulates 

10.  The  postulates  of  geometry  are  fundamental  agree- 
ments or  conventions  concerning  the  starting  point  and 
scope  of  the  science. 


PBIMARY  DEFINITIONS  3 

11.  Postulate  of  space-dimensions.  It  is  commonly  agreed 
that  ordinary  geometry  shall  treat  only  of  a  space  of  three 
dimensions. 

We  cannot,  however,  assert  that  a  space  of  four  dimensions  could 
not  exist  under  any  conditions.  We  are  not  able  to  form  a  mental 
picture  of  such  a  space,  but  it  does  not  follow  that  no  one  will  ever  be 
able  to  form  such  a  picture. 

12.  Postulate  of  figure-transference.  It  is  also  commonly 
agreed  that  ordinary  geometry  shall  consider  only  a  space  in 
which  figures  can  be  transferred  in  thought  from  one  posi- 
tion to  another  without  further  change. 

There  is  a  branch  of  higher  geometry  which  considers  the  possible 
existence  of  a  space  in  which  figures  are  not  transferable  without 
change.     (See  Art.  34. ) 

Besides  the  two  postulates  just  stated,  other  postulates 
will  be  introduced  in  due  course. 

Primary  Definitions 

13.  In  geometry  a  definition  is  a  statement  of  what  is 
to  be  regarded  as  the  fundamental  property  of  a  certain  class 
of  figures,  sufficient  to  distinguish  the  class,  and  also  suffi- 
cient to  furnish  a  starting  point  for  deriving  other  proper- 
ties by  logical  inference. 

The  definitions  will  be  introduced  wherever  occasion 
arises.  The  name  to  be  applied  to  the  class  of  figures  so 
defined  will  be  italicized  when  used  for  the  first  time  in 
the  definition. 

14.  Superposable  figures.  Equal  figures.  If  two  figures  are 
such  that  they  can,  by  transference,  be  so  applied  to  each 
other  that  every  point  of  one  falls  on  some  point  of  the 
other,  point  for  point,  the  two  figures  when  so  applied  are 
said  to  be  coincident,  or  to  be  superposed.  Figures  that 
are  capable  of  superposition  are  said  to  be  superposable. 

Superposable  figures  are  also  said  to  be  equal  to  each  other. 
Thus  the  phrase  "equal  figures''  will  always  have  the 
same  meaning  as  "superposable  figures." 


4  INTRODUCTION 

Straightness  as  a  quality  of  certain  lines. 

15.  The  fundamental  meaning  of  straightness  as  a  geomet- 
ric concept  is  to  be  obtained  by  idealizing  our  experience. 

It  is  well  known  that  the  practical  straightness  of  two 
rulers,  for  instance,  is  tested  by  observing  whether  their 
edges  seem  to  fit  each  other,  no  matter  how  they  may  be 
moved  or  turned.  If  no  want  of  coincidence  could  be  revealed 
by  any  microscope,  however  powerful,  both  edges  would 
have  the  ideal  quality  of  straightness  in  the  geometric  sense. 

Accordingly  the  notion  of  straightness  as  possessed  by 
certain  ideal  lines  is  embodied  in  the  following  definition. 

16.  Straight  lines  are  lines  of  unlimited  extent  such 
that  any  portion  of  any  of  them  will  coincide  with  any  other 
portion  of  any  of  them,  however  applied,  if  the  extremities 
of  the  two  portions  coincide. 

17.  It  follows  from  this  definition  that  if  two  straight 
lines  pass  through  the  same  two  points,  the  lines  coincide, 
and  may  then  be  regarded  as  the  same  line. 

This  may  be  conveniently  expressed  thus : 
Only  one  straight  line  can  pass  through  the  same  two 
points. 

18.  Broken  line.  Curved  line.  A  line  composed  of  parts 
of  different  straight  lines  is  called  a  broken  line.  A  line 
of  which  no  part  is  straight  is  called  a  curved  line,  or  curve. 

Flatness  as  a  quality  of  certain  surfaces. 

19.  Again,  it  is  well  known  that  the  practical  flatness  of 
a  surface  is  tested  by  observing  whether  a  straightedge  fits 
it,  however  placed  on  the  surface. 

Accordingly  the  notion  of  perfect  flatness  as  possessed  by 
certain  ideal  surfaces  is  embodied  in  the  following  definition. 

20.  A  flat  surface,  or  plane,  is  a  surface  of  unlimited 
extent  such  that  a  straight  line  passing  through  any  two  of 
its  points  lies  wholly  in  the  surface. 


BOUNDARIES   OF  SEPARATION  5 

21.  It  follows  from  this  definition,  and  from  the  definition 
of  a  straight  line,  that  plane  surfaces  are  such  that  any 
portion  of  any  of  them  will  coincide  with  any  other  portion 
of  any  of  them,  however  applied,  if  the  boundaries  of  the 
two  portions  coincide.  For  the  straight  line  passing  through 
any  two  points  of  the  common  boundary  must  lie  in  both 
planes. 

22.  Figures  formed  by  points  and  lines  traced  on  a  plane 
surface  are  called  plane  ftgures. 

The  study  of  plane  figures  is  called  plane  geometry,  or 
geometry  of  two  dimensions.  The  consideration  of  all  other 
figures  belongs  to  solid  geometry. 

Boundaries  of  Separation 

23.  Two  portions  of  a  plane  that  have  a  common  bound- 
ing line  are  said  to  be  separated  by  that  boundary  if  every 
line  passing  from  a  point  of  one  portion  to  a  point  of  the 
other,  and  not  passing  out  of  both  portions,  passes  through 
some  point  of  the  common  bounding  line. 

Two  portions  of  space  that  have  a  common  bounding  sur- 
face are  said  to  be  separated  by  that  boundary  if  every 
line  passing  from  a  point  of  one  portion  to  a  point  of  the 
other,  and  not  passing  out  of  both  portions,  passes  through 
some  point  of  the  common  bounding  surface. 

24.  Postulate  of  separation.  Let  it  be  granted  that  an  un- 
limited straight  line  on  a  plane  surface  divides  the  whole 
plane  into  two  portions  that  are  separated  from  each  other 
by  the  straight  line. 

25.  Each  of  the  two  parts  into  which  a  plane  is  divided 
by  an  unlimited  straight  line  is  called  a  half  plane.  Two 
figures  in  the  same  half  plane  are  said  to  be  at  the  same 
side  of  the  straight  line.  Two  figures,  one  in  each  half 
plane,  are  said  to  be  at  opposite  sides  of  the  line ;  and  the 
line  is  said  to  pass  between  the  two  figures. 


6  INTRODUCTION 

26.  A  straight  line  terminating  at  a  point  and  extending 
indefinitely  the  other  way  is  called  an  indefinite  half  line. 
Thus  any  assigned  point  on  an  unlimited  straight  line 
divides  the  line  into  two  indefinite  half  lines.  Two  points 
on  the  same  half  line  are  said  to  be  at  the  same  side  of  the 
assigned  point.  Two  points,  one  on  each  half  line,  are  said 
to  be  at  opposite  sides  of  the  point  of  separation,  and  the 
latter  point  is  said  to  be  between  the  two  former. 

CLOSED    FIGURES 

27.  A  line  on  a  plane  surface  is  said  to  be  closed  if  it 
separates  a  finite  portion  of  the  plane  from  the  remaining 
indefinite  portion. 

A  surface  is  said  to  be  closed  if  it  separates  a  finite  por- 
tion of  space  from  the  remaining  indefinite  portion. 

In  both  cases  the  finite  portion  is  said  to  be  inclosed  by 
the  boundary.  All  points  of  the  finite  portion,  not  on  the 
boundary,  are  said  to  be  within  the  figure ;  and  all  other 
points,  not  on  the  boundary,  are  said  to  be  without  the  figure. 

Thus  any  line  passing  from  any  point  within  to  any  point 
without  a  closed  figure  passes  through  some  point  of  the 
boundary. 

Hence  an  unlimited  straight  line  passing  through  any 
point  within  the  closed  figure  passes  through  at  least  two 
points  of  the  boundary. 

Roundness  as  a  quality  of  certain  closed  figures. 

28.  The  notion  of  roundness  as  applied  to  certain  figures 
is  embodied  in  the  following  definitions : 

A  circle  is  a  plane  closed  line  such  that  all  straight  lines 
joining  any  point  on  this  line  to  a  certain  point  within  the 
figure  are  equal.    This  point  is  called  the  center  of  the  circle. 

A  sphere  is  a  closed  surface  such  that  all  straight  lines 
joining  any  point  on  the  surface  to  a  certain  point  within 
the  figure  are  equal.  This  point  is  called  the  center  of  the 
sphere. 


THE  CONSTRUCTION  POSTULATES  7 

In  the  circle  or  the  sphere  a  line  joining  the  center  to  any 
point  on  the  boundary  is  called  a  radius. 

The  property  of  having  equal  radii  from  a  certain  point 
to  the  boundary  is  called  roundness. 

Thus  a  circle  is  a  round  plane  curve;  and  a  sphere  is  a 
round  surface.     A  portion  of  a  circle  is  called  an  arc. 

The  Construction  Postulates 

29.  The  conventions  in  elementary  geometry  with  regard 
to  the  recognized  ways  of  constructing  figures  on  a  plane 
surface  are  expressed  in  the  following  construction  postu- 
lates : 

Let  it  be  granted : 

1.  That  a  straight  line  may  be  drawn  from  any  one  point 
to  any  other. 

2.  That  a  terminated  straight  line  may  be  prolonged  in- 
definitely. 

3.  That  a  circle  may  be  described  on  a  plane  surface  with 
any  point  of  the  plane  as  center,  and  with  a  radius  equal  to 
any  finite  straight  line. 

30.  We  are  thus  to  be  allowed  the  use  of  the  straightedge 
for  drawing  and  prolonging  lines  on  a  plane  surface ;  and 
also  the  use  of  the  compasses  for  describing  circles,  and  for 
transferring  portions  of  straight  lines. 

The  lines  so  drawn  will  not  indeed  be  true  geometrical 
lines,  however  finely  they  may  be  traced.  One  of  the  pur- 
poses of  the  postulates  is  to  make  an  agreement  by  which 
the  lines  so  traced  shall  be  regarded  as  representing  true 
lines.  They  will  be  supposed  to  have  no  irregularities,  and 
to  cover  no  portion  of  the  surface,  being  thought  of  as  mere 
boundaries. 

31.  Besides  this  positive  use  of  the  construction  postulates, 
they  have  also  a  negative  or  restrictive  use.  No  construction 
is  to  be  allowed  in  elementary  plane  geometry  which  cannot 
be  performed  by  a  combination  of  the  primary  constructions. 

MCM.  ELEM.  GEOM. — 2 


8  INTRODUCTION 

On  the  Fundamental  Conventions 

32.  Postulates  of  existence.  Besides  the  postulates  that 
have  been  formally  stated  above,  or  that  may  hereafter  be 
introduced,  there  are  certain  other  fundamental  conventions, 
which  should  be  noticed.  The  foregoing  definitions  contain 
implied  agreements  that  the  thing  defined  shall  be  regarded 
as  existing.  These  implied  agreements  are  called  postulates 
of  existence.  Thus  by  means  of  the  definitions  we  have 
postulated  the  existence  of  points,  lines,  and  surfaces,  in 
general,  and  also  the  existence  of  particular  lines  and  sur- 
faces having  the  respective  qualities  of  straightness,  flatness, 
closedness,  and  roundness. 

33.  Twofold  purpose  of  the  postulates.  It  should  be  observed 
that  our  space  postulates,  whether  expressed  or  implied,  are 
not  of  an  arbitrary  nature,  for  they  are  the  outcome  of  our 
actual  space-experience.  The  postulates,  however,  go  be- 
yond our  experience  in  two  ways.  In  the  first  place  they 
raise  to  ideal  exactness  our  ordinary  perceptions  of  space, 
which  are  more  or  less  crude.  Again,  by  means  of  the  postu- 
lates, we  extend  to  the  space  outside  of  our  experience  the 
primary  notions  suggested  by  our  perception  of  the  limited 
portion  of  space  that  we  inhabit. 

34.  The  postulates  as  defining  £uclidean  space.  A  space  that 
fulfills  the  conditions  embodied  in  the  postulates  and  primary  definitions 
is  called  a  Euclidean  space  after  the  name  of  Euclid,  who  wrote  the  first 
systematic  treatise  on  geometry.  We  can  never  be  absolutely  certain, 
at  least  with  our  present  mode  of  perception,  that  our  space  is  of  the 
ideal  Euclidean  character,  but  there  is  no  doubt  that,  for  all  human 
needs,  it  may  be  regarded  as  accurately  Euclidean. 

A  perfect  system  of  postulates  should  embody  the  primary  notions 
that  are  necessary  and  sufficient  to  distinguish  Euclidean  space  from 
other  kinds  of  space,  and  to  furnish  a  starting  point  from  which  all  its 
properties  could  be  derived  by  a  chain  of  reasoning  without  further 
resort  to  experience.  Euclid  and  the  ancient  geometers  did  not  give 
close  attention  to  the  necessity  and  sufficiency  of  their  system  of  con- 
ventions.    They  silently  took  for  granted  certain  things  that  do  not 


PBIMARY  MAGNITUDE  RELATIONS  9 

follow  from  previously  accepted  principles ;  and  some  of  their  fimda- 
mental  conventions  are  not  independent  of  each  other.  Modern  geom- 
eters are  not  yet  entirely  agreed  on  a  complete  system  of  mutually 
independent  postulates  for  Euclidean  space,  and  the  full  discussion  of 
this  question  goes  beyond  the  limits  set  for  elementary  geometry. 

Several  different  systems  of  non-Euclidean  geometry  have  been 
studied,  each  of  which  dispenses  with  one  or  more  of  the  characteristic 
properties  of  Euclidean  space.  The  most  celebrated  of  these  systems 
dispenses  with  the  'postulate  of  parallels,'  which  will  be  introduced  in 
the  proper  place.  It  should  be  observed  that  if  only  the  '  postulate  of 
dimensions '  is  dispensed  with,  the  space  is  still  Euclidean  in  character, 
for  a  Euclidean  three-dimensional  space  may  be  regarded  as  existing  in 
a  Euclidean  space  of  four  or  more  dimensions,  just  as  a  two-dimensional 
space  exists  in  a  three-dimensional  one.  A  Euclidean  space  of  more 
than  three  dimensions  is  called  a  Euclidean  hyper-space. 

Primary  Magnitude  Relations 

35.  Definitions.  A  magnitude  is  anything  that  is  divis- 
ible into  parts. 

A  magnitude  is  said  to  be  the  sum  of  all  its  parts. 

A  magnitude  is  said  to  be  greater  than  any  part  of  it, 
and  also  greater  than  any  other  magnitude  which  is  super- 
posable  on  a  part  of  it. 

In  the  same  way  a  part  is  said  to  be  less  than  the  whole  ; 
and  any  magnitude  that  is  superposable  on  a  part  is  also 
said  to  be  less  than  the  whole. 

If  a  magnitude  is  divided  into  any  two  parts,  either  part 
is  said  to  be  the  difference  of  the  whole  and  the  other  part. 

If  a  magnitude  is  divided  into  two  superposable  parts,  each 
part  is  said  to  be  half  of  the  whole,  and  the  whole  is  said  to 
be  double  of  either  part. 

36.  It  will  next  be  shown  how  to  construct  the  sum  and 
difference  of  certain  magnitudes  ;  and  some  of  the  terms  in- 
troduced above  will  receive  more  detailed  definition  in  con- 
nection with  the  special  magnitudes  to  which  they  are  applied. 

The  simplest  magnitudes  are  straight  lines  and  plane 
angles.    These  will  be  considered  in  the  next  section. 


PLANE   GEOMETKT 


BOOK  I.  — RECTILINEAR  FIGURES 

LINE-SEGMENTS  AND  ANGLES 
Definitions  concerning  Line-segments 

1.  Any  portion  of  a  straight  line  is  called  a  line-segment. 
It  is  usually  designated  by  two  letters,  one  placed  at  each 
extremity. 

Any  two  segments  of  the  same  straight  line  are  called 
collinear  segments. 

If  two  collinear  segments  have  a  common  extremity,  and 
are  at  opposite  sides  of  this  common  point,  they  are  called 
adjacent  collinear  segments;  and  the  segment  which  is 
composed  of  them  is  called  their  sum. 

Several  collinear  segments  are  said  to  be  consecutive  when 
the  second  is  adjacent  to  the  first,  the  third  to  the  second,  and 
so  on  without  overlapping;  and  the  whole  segment  com- 
posed of  them  is  called  their  suw/. 

For  instance,   the   segment 

AE  is  the  sum  of  the  segments    -^ ^ g — J -^ — 

AB,  BCy  CD,  DE. 

The  segment  AE  is  also  the  sum  of  the  segments  AB,  BD,  DE ; 
and  of  AC,  CE. 

Again,  the  sum  of  any  line-segments,  however  situated,  is 
the  segment  obtained  by  transferring  them  so  as  to  be  con- 
secutive collinear  segments,  and  then  taking  their  sum  by  the 
preceding  definition. 

10 


p 

Q 

A 

B 

C 

LINE-SEGMENTS  AND  ANGLES  11 

Comparison  of  line-segments. 

2.  Two  given  line-segments  may  be  compared  by  trans- 
ferring one  or  both  so  that  they  may  become  collinear,  have 
a  common  extremity,  and  lie  on  the  same  side  of  this 
extremity. 

E.g.,  to  compare  the  segments  PQ  and  RS,  take  an  indefinite 
line,  and  transfer  PQ  to  the  position  AC,  and  RS  to  the  posi- 
tion AB.  Then  they  have  a  com- 
mon extremity  A.  If  the  other 
two  extremities  B  and  C  happen 
to  coincide,  the  segments  PQ  and 
RS  are  equal.  If  these  extremi-  s" 
ties  do  not  coincide,  and  if  the 

three  points  are  in  the  order  A,  B,  C,  then  the  segment  ^C  is 
the  sum  of  AB  and  another  segment  BC. 

In  this  case  AC  is  said  to  be  greater  than  its  part  AB,  and 
the  latter  is  said  to  be  less  than  the  former.  Accordingly, 
PQ  is  said  to  be  greater  than  RS,  and  RS  is  said  to  be  less 
than  PQ  (Introd.  35). 

The  other  segment  BC  which  when  added  to  the  less  pro- 
duces the  greater  is  called  the  difference  of  the  two  given 
segments.  It  is  also  called  the  excess  of  the  greater  over 
the  less,  or  the  remainder  obtained  by  subtraxjting  the  less 
from  the  greater. 

3.  The  sum  of  two  equal  line-segments  is  called  the 
double  of  either  of  them ;  and  each  of  the  equal  segments 
is  said  to  be  half  of  the  whole  segment  which  is  composed 
of  them. 

4.  Hereafter  the  word  line  when  used  without  a  quali- 
fying word  will  mean  straight  line. 

Sometimes  a  line-segment  will  be  called  a  line  when 
there  is  no  possibility  of  mistaking  it  for  an  indefinite  line. 

The  figure  which  next  presents  itself  is  that  formed  by 
two  lines  terminated  at  the  same  point. 


12  PLANE  GEOMETRY — BOOK  I 

Definitions  concerning  Angles 

5.  An  angle  is  the  figure  formed  by  two  indefinite  half 
lines  issuing  from  the  same  point.  This  point  is  called  the 
vertex  of  the  angle,  and  the  half  lines  are  called  its  sides. 

An  angle  is  usually  designated  by  three  letters,  the  middle 
one  being  placed  at  the  vertex  and  the  other  two 
on  the  sides ;  thus  the  angle  of  the  straight  lines 
ABj  AC  \%  called  the  angle  BAC.  When,  however, 
there  is  no  other  angle  having  the  same  vertex, 
the  letter  at  the  vertex  is  a  sufficient  designation. 

6.  A  useful  notion  of  an  angle  may  be  obtained  by  the 
conception  of  a  revolving  line. 

In  the  angle  AOB,  imagine  a  line  at  first  to  coincide  with 
OA  and  then  to  revolve  about  the  point  0 
(that  is,  to  coincide  in  succession  with  dif- 
ferent lines  passing  through  O)  until  it 
arrives  in  the  position  OB.  The  revolving 
line  is  then  said  to  have  turned  through 
the  angle  AOB. 

As  there  are  two  ways  of  turning  a  line  from  the  position 
OA  to  the  position  05,  there  are  two  angles  AOB  formed  by 
the  same  two  half  lines.  These  are  said  to  be  conjunct 
angles. 

7.  Two  angles  are  said  to  be  equal  (in  accordance  with 
the  definition  of  equal  figures  in  general)  when  either  angle 
may  be  transferred  so  as  to  coincide  with  the  other,  i.e.  so 
that  their  sides  may  be  coincident,  and  so  that  the  two 
angles  in  question  can  then  be  turned  through  at  the  same 
time  by  the  revolving  half  line. 

When  two  angles  are  in  coincidence,  their  conjunct  angles 
are  also  in  coincidence. 

8.  Two  angles  that  have  the  same  vertex  with  one  side 
common,  and  are  situated  at  opposite  sides  of  this  common 


LINE-SEGMENTS  AND  ANGLES  13 

line,  are  called   adjacent  angles,    and  the  whole  angle 

formed  by  the  two  extreme  lines,  of  which 

these  two   angles   are  parts,  is  called  the 

sum  of  the  two  adjacent   angles.     Thus, 

the  sum  of  the  angles  AOB  and  BOC  is  the 

angle  AOC. 

^  A. 

9.  Several  angles  are  said  to  be  adjacent  in  succession 
when  they  have  a  common  vertex  and  are  such  that  the 
second  is  adjacent  with  the  first,  the 
third  with  the  second,  and  so  on  with- 
out overlapping;  and  the  whole  angle 
formed  by  the  two  extreme  lines,  of 
which  these  angles  are  parts,  is  called 
the  sum  of  the  several  angles.  For 
instance,  the  angle  AOE  is  the  sum  of  the  angles  AOB,  BOC, 
COD,  and  DOE.  It  is  likewise  the  sum  of  the  angles  AOCy 
COD,  and  DOE. 

10.  Again,  the  sum  of  several  angles  not  adjacent  in  suc- 
cession is  the  angle  obtained  by  transferring  them  so  as  to 
be  adjacent  in  succession,  and  then  taking  their  sum  accord- 
ing to  the  preceding  definition.     This  process  is  called  the 


addition  of  angles,  and  the  given  angles  are  then  said  to 
be  added  or  summed.  It  may  be  stated  as  the  process  of 
letting  a  revolving  line  turn  successively  through  angles 
equal  to  the  given  angles  (such  as  1,  2,  3)  ;  the  wliole  angle 
thus  turned  through  being  the  required  sum  of  the  separate 
angles. 


14  PLANE  GEOMETE  Y  —  BOOK  I 

Comparison  of  angles. 

11.  Two  angles  are  compared  in  regard  to  magnitude  by 
transferring  one  or  both  so  that  they  may  have  the  same 
vertex,  a  common  side,  and  lie  at  the  same  side  of  this  com- 
mon line.  If  the  other  two  sides  happen  to  coincide,  the 
angles  are  equal.  If  these  sides  do  not  coincide,  one  of 
the  given  angles  is  equal  to  the  sum  of  the  other  given  angle 
and  a  third  angle.  The  first  given  angle  is  then  said  to  be 
greater  than  the  other,  and  the  latter  is  said  to  be  less  than 
the  former  (Introd.  35). 

12.  The  third  angle,  mentioned  above  (11),  which  when 
added  to  the  less  produces  the  greater,  is  called  the  differ- 
ence of  the  two  given  angles.  It  is  called  also  the  excess 
of  the  greater  over  the  less,  or  the  remainder  obtained  by 
taking  the  less  away  from  the  greater. 

13.  One  angle  is  said  to  be  the  double  of  another,  if  it 
is  the  sum  of  two  angles  each  equal  to  the  other ;  and  the 
latter  angle  is  called  the  half  of  the  former. 

14.  It  will  be  seen  from  the  above  definitions  of  the  words 
equal,  sum,  difference,  greater,  less,  double,  half,  when 
applied  to  angles,  that  in  comparing  the  magnitude  of  differ- 
ent angles  nothing  is  said  about  the  magnitude  of  their 
sides.  In  fact,  the  sides  of  an  angle  may  always  be  thought 
of  as  indefinitely  prolonged. 

Species  of  angles. 

15.  When  the  revolving  half  line  turns  from  the  position 
OA  into  the  position  OA',  the  prolongation  of  OA^  it  is  then 
said  to  have  turned  through  a  straight  angle. 


a: 


Thus,  an  angle  whose  sides  are  in  the  same  straight  line 
at  opposite  sides  of  the  vertex,  is  a  straight  angle. 


LINE-SEGMENTS  AND  ANGLES 


15 


16.  If  this  revolving  line  turns  through  another  straight 
angle,  from  OA'  to  OA,  so  as  to  complete  a  revolution,  the 
angle  turned  through  is  called  a  perigon. 


A' K^ A 

17.   The  half  of  a  straight  angle  is  called  a  right  angle. 
B 

L. 


18.  An  angle  less  than  a  right  angle  is  called  an  acute 
angle. 

19.  An  angle  greater  than  a  right  angle  and  less  than  a 
straight  angle  is  called  an  obtuse  angle. 

20.  An  angle  less  than  a  straight  angle  is  called  a  con- 
cave angle.  Two  concave  angles  are  said  to  be  of  the 
same  species  when  they  are  both  acute,  both  right,  or  both 
obtuse. 


21.  An  angle  greater  than  a  straight  angle  and  less  than 
a  perigon  is  called  a  convex  angle. 

22.  The  definitions  above  given  and  illustrated  (1-21) 
form  the  basis  of  the  statements  in  the  next  section.  Fur- 
ther definitions  will  be  introduced  as  occasion  requires. 


16  PLAJI^E  GEOMETRY  —  BOOK  I 

Axioms  concerning  Lines  and  Angles* 

23.  An  axiom  is  a  general  statement  whose  truth  can  be 
immediately  inferred  from  the  definitions  of  the  terms  used. 

It  is  convenient  for  purpose  of  reference  to  designate 
specially  by  the  term  axiom  some  fundamental  statements, 
relating  to  the  equality  and  inequality  of  magnitudes,  whose 
truth  can  be  inferred  directly  from  the  above  definitions. 

We  apply  these  axioms  only  to  those  magnitudes  for 
which  the  appropriate  methods  of  comparison  have  been 
already  explained. 

When  other  kinds  of  magnitude  are  introduced,  and  when 
all  the  terms  employed  receive  precise  definitions  as  applied 
to  such  magnitudes,  then  the  appropriate  axioms  will  be 
stated,  and  their  truth  inferred  from  the  definitions. 

The  first  seven  of  the  following  axioms  relate  to  the 
equality  of  magnitudes,  the  remaining  seven  to  inequality. 

24.  Ax.  1.  Magnitudes  which  are  equal  to  the  same  mag- 
nitude are  equal  to  each  other. 

25.  Ax.  2.  If  equal  magnitudes  are  added  respectively  to 
equal  magnitudes,  the  sums  are  equal. 

26.  Ax.  3.  If  equal  magnitudes  are  subtracted  respectively 
from  equal  magnitudes,  the  differences  are  equal. 

27.  Ax.  4.   The  doubles  of  equal  magnitudes  are  equal. 

28.  Ax.  5.   The  halves  of  equal  magnitudes  are  equal. 

29.  (a)  Ax.  6.  The  sum  of  several  magnitudes  taken  in 
any  order  is  equal  to  their  sum  taken  in  any  other  order. 

(h)  Ax.  7.  The  double  of  the  sum  of  two  magnitudes  is 
equal  to  the  sum  of  their  doubles. 

*  The  student  need  not  dwell  on  Arts.  23-40  at  first  reading,  but 
should  refer  back  to  them  when  necessary. 


LINE-SEGMENTS  AND  ANGLES  17 

30.  Ax.  8.  If  one  magnitude  is  equal  to  or  greater  than 
a  second  magnitude,  and  the  second  greater  than  a  third, 
then  the  first  is  greater  than  the  third,  and  also  greater  than 
any  magnitude  equal  to  the  third. 

31.  (a)  Ax.  9.  If  equal  magnitudes  are  added  respectively 
to  unequal  magnitudes,  the  sums  are  unequal,  the  greater  sum 
arising  from  the  addition  of  the  greater  magnitude. 

(b)  Ax.  10.  If  equal  magnitudes  are  subtracted  respec- 
tively from  unequal  magnitudes,  the  differences  are  unequal, 
the  greater  difference  being  part  of  the  greater  magni- 
tude. 

(c)  Ax.  11.  If  unequal  magnitudes  are  subtracted  respec- 
tively from  equal  magnitudes,  the  differences  are  unequal, 
the  greater  difference  arising  from  the  subtraction  of  the 
less  magnitude. 

32.  (a)  Ax.  12.  If  unequal  magnitudes  are  added  to  un- 
equal magnitudes,  the  sum  of  the  two  greater  magnitudes 
is  greater  than  the  sum  of  the  two  less. 

(6)  Ax.  13.  If  two  magnitudes  are  unequal,  the  double  of 
the  greater  is  greater  than  the  double  of  the  less. 

(c)  Ax.  14.  If  two  magnitudes  are  unequal,  the  half  of 
the  greater  is  greater  than  the  half  of  the  less. 

33.  The  method  by  which  the  truth  of  any  one  of  these 
axioms  is  derived  from  the  definitions  is  called  hmnediate 
inference,  because  the  whole  process  consists  of  only  a 
single  step. 

We  have  next  to  show  how  a  new  geometric  truth  may  be 
derived  from  definitions,  postulates,  axioms,  or  other  accepted 
facts,  by  a  process  of  mediate  inference,  that  is,  by  a  series 
of  steps  intermediate  between  the  accepted  facts  and  the 
new  truth. 

It  is  here  that  geometry,  like  other  sciences,  calls  in  the 
aid  of  logic,  the  science  of  reasoning. 


18  PLANE  GEOMETRY — BOOK  I 

Some  Logical  Terms  used  in  Geometry 

34.  A  theorem  is  a  statement  enunciating  a  fact  whose 
truth  can  be  inferred  from  other  statements  previously 
accepted  as  true. 

The  enunciation  of  a  theorem  consists  of  two  parts :  the 
hypothesis,  or  formal  statement  of  the  conditions ;  and  the 
conclusion,  or  that  which  is  asserted  to  follow  necessarily 
from  the  hypothesis. 

35.  The  process  by  which  it  is  made  clear,  step  by  step, 
that  the  conclusion  must  be  true  if  the  hypothesis  is  true  is 
called  the  clemonsti^ation  or  proof  of  the  theorem.  Each 
step  in  the  demonstration  must  be  authorized  by  something 
previously  accepted  as  true. 

36.  A  corollary  to  a  theorem  is  a  statement  whose  truth 
follows  at  once  from  the  truth  of  the  theorem,  or  which  can 
be  proved  by  a  similar  course  of  reasoning. 

A  corollary  to  a  theorem  may  be  used  (like  the  theorem 
itself)  in  the  proof  of  a  subsequent  theorem  or  corollary. 

37.  In  geometry  a  theorem  relates  to  a  certain  kind  of 
figure,  and  asserts  that  if  the  figure  possesses  a  certain  prop- 
erty stated  in  the  hypothesis,  then  it  must  also  possess  a 
certain  other  property  stated  in  the  conclusion. 

The  theorem  will  first  be  stated  in  general  terms  so  as 
to  apply  to  a  whole  class  of  figures  possessing  a  certain  com- 
mon property.  This  statement  is  called  the  general  enun- 
ciation, or  simply  the  enunciation,  of  the  theorem. 

38.  For  convenience  a  single  representative  figure  will 
be  drawn ;  and  the  assumed  hypothesis,  with  the  conclusion 
to  be  derived  from  it,  will  both  be  restated  with  special 
reference  to  the  particular  figure  so  drawn.  Such  restate- 
ment is  called  the  special  enunciation  of  the  theorem. 

The  successive  steps  of  the  demonstration  will  also  be 
explained  with  regard  to  this  figure,  the  authority  for  each 
advance  being  quoted  until  the  conclusion  is  reached. 


LINE-SEGMENTS  AND  ANGLES  19 

39.  Typical  fonn  of  geometric  theorem.  The  hypothesis 
of  a  geometric  theorem  can  usually  be  put  in  the  type-form 
'^A  is  5,"  and  the  conclusion  in  the  form  "  C  is  D." 

Each  of  these,  considered  separately,  is  called  a  simple 
statement,  or  a  simple  proposition. 

When  two  simple  propositions  are  brought  together,  the 
first  being  preceded  by  the  word  if  or  the  word  when,  and 
the  second  by  the  word  then,  they  are  said  to  form  a 
hypothetical  proposition. 

In  the  type-form  of  hypothetical  proposition, 
"  if  A  is  B,  then  C  is  D," 
the  assertion  is  that  the  second  statement  (the  conclusion) 
is  a  necessary  consequence  of  the  first  (the  hypothesis),  so 
that  any  one  who  agrees  to  the  first  must  also  accept  the 
second.  The  hypothesis  is  sometimes  called  the  antecedent, 
and  the  conclusion  the  consequent. 

It  is  to  be  understood  that  when  A  and  C  stand  for  plural 
nouns,  the  plural  verb  will  be  used. 

All  geometric  theorems  are  of  the  above  tjrpe-form,  or  can 
be  put  in  this  form. 

For  instance,  "  The  halves  of  equal  angles  are  equal "  can 
be  stated  in  the  hypothetical  form, 

"  If  two  angles  are  equal,  then  their  halves  are  equal." 
Sometimes  it  is  more  convenient  not  to  use  the  hypothetical 
form  of  statement  in  the  general  enunciation  of  a  theorem ; 
but  the  hypothetical  form  (or  something  equivalent  to  it)  is 
always  used  in  the  special  enunciation. 

For  instance,  in  Theorem  2,  below,  the  general  enunciation, 
"All  right  angles  are  equal,"  is  an  abbreviation  of  the  hypo- 
thetical proposition, 
"  If  any  two  angles  are  right  angles,  then  they  are  equal." 

In  the  special  enunciation  the  word  if  is  replaced  by  the 
word  let;  and  the  word  then  by  the  words  to  prove: 

"Let  AOB  and  A^O'B^  (referring  to  the  figure)  be  any  two 
right  angles.     To  prove  that  they  are  equal." 


20  PLANE  GEOMETRY — BOOK  I 

40.  Method  of   arrangement.     The  general  enunciation 

is  placed  first,  and  printed  in  italics. 

Next  in  order  is  the  special  enunciation,  which  consists 
of  two  parts :  (1)  the  special  statement  of  the  hypothesis, 
introduced  by  the  word  let,  and  preceding  the  figure  to 
which  it  refers ;  (2)  the  special  statement  of  the  conclusion 
to  be  demonstrated,  introduced  by  the  words  to  prove,  fol- 
lowing immediately  after  the  figure. 

Any  construction  lines  that  may  aid  in  the  proof  are  next 
indicated,  and  are  usually  dotted  in  the  drawing  to  distin- 
guish them  from  the  lines  mentioned  in  the  hypothesis, 
which  are  drawn  full. 

The  successive  steps  in  the  demonstration  leading  from 
hypothesis  to  conclusion  are  then  made  clear  with  reference 
to  the  figure  so  drawn,  the  previous  authority  for  each 
step  being  quoted,  or  referred  to.  The  authority  may  con- 
sist in  the  hypothesis,  a  definition,  a  postulate,  an  axiom, 
a  previous  theorem,  or  corollary. 

Theorems  concerning  Angles 

41.  Theorem  1.    All  straight  angles  are  equal. 

Let  the  indefinite  lines  OA,  OB  be  the  sides  of  a  straight 
angle  whose  vertex  is  o ;  and  let  0'A\  O'b'  be  the  sides  of 
a  straight  angle  whose  vertex  is  0'. 

:t-4— i = ^•W— ' 

B  0  A  B'  0'  A' 

To  prove  that  the  straight  angle  formed  by  the  lines  OA 
and  OB  is  equal  to  the  straight  angle  formed  by  the  lines 
O'A'  and  0'B\ 

Because  the  first-mentioned  angle  is  a  straight  angle, 
therefore  BO  and  OA  are  in  the  same  straight  line  [by  the 
definition  of  a  straight  angle  (15)]. 

Similarly  B^O^  and  O'^'  are  in  the  same  straight  line. 

Then  the  line  BOA  can  be  superposed  on  the  line  B^O^A^ 


LINE-SEGMENTS  AND  ANGLES  21 

SO  that  the  point  0  falls  on  0',  by  the  postulate  of  trans- 
ference (Introd.  12),  and  the  definition  of  straight  lines 
(Introd.  16).  The  two  straight  angles  will  then  have  their 
vertices  and  sides  in  exact  coincidence.  Therefore  they  are 
equal  [by  the  definition  of  equal  angles  (7)]. 

42.   Theorem  2.     All  right  angles  are  equal. 
Let  AOB,  A'O'B'  be  any  two  right  angles. 
B  B' 


0  A  0'  A' 

To  prove  that  these  angles  are  equal. 
Every  right  angle  is  half  a  straight  angle  (17,  def.). 
Now  all  straight  angles   are  equal   (theor.  1) ;    and  the 
halves  of  equal  angles  are  equal  (ax.  5,  28). 
Therefore  all  right  angles  are  equal. 

43.  Cor.  Tlie  sum  of  any  two  ^  right  angles  is  equal  to  a 
straight  angle. 

44.  Theorem  3.     All  perigons  are  equal. 

Let  a  straight  line  terminated  at  0  and  indefinitely  ex- 
tended toward  A  revolve  about  its  extremity  0  from  the 
position  OA  through  a  perigon  into  the  position  OA  again. 
Similarly  let  a  line  revolve  about  O'  from  the  position  OU' 
through  a  perigon  into  the  position  O'J'  again. 


l-Tt- 


To  prove  that  these  perigons  are  equal. 
Each  perigon  is  the  sum  of  two  straight  angles  (15,  16). 
Now  all  straight  angles  are  equal  (theor.  1) ;  and  the  sums 
of  equal  angles  are  equal  (25,  ax.  2). 
Therefore  the  perigons  are  equal. 


22  PLANE  GEOMETRY  —  BOOK  I 

45.  Cor.  The  sum  of  any  four  right  angles  is  equM  to  a 
perigon. 

COMPLEMENTS,    SUPPLEMENTS,    ETC. 

46.  Definitions.  If  two  lines  form  a  right  angle  each  is 
said  to  be  perpendicular  to  the  other. 

If  two  perpendicular  lines  are  prolonged  through  their  in- 
tersection, they  divide  the  perigon  into  four  equal  parts  (17). 

47.  When  the  sum  of  two  angles  is  a  right  angle,  each  is 
called  the  complement  of  the  other. 

When  the  sum  of  two  angles  is  a  straight  angle,  each  is 
called  the  supplement  of  the  other. 

When  the  sum  of  two  angles  is  a  perigon,  each  is  called 
the  conjunct  of  the  other. 

From  the  definitions  the  two  following  statements  are 
immediate  inferences. 

48.  Two  adjacent  angles  are  complemental  if  their  ex- 
terior sides  form  a  right  angle  which  includes  the  two  angles. 


49.  Two  adjacent  angles  are  supplemental  if  their  exterior 
sides  form  a  straight  angle. 

50.  Theorem  4.  Covzplem^nts  of  equal  angles  are 
equal. 

Proof.  The  complement  of  each  angle  is  obtained  by- 
subtracting  it  from  a  right  angle  (def.,  47). 

Now  all  right  angles  are  equal  (theor.  2,  42)  ;  and  if  equal 
angles  are  subtracted  from  equal  angles,  the  remainders  are 
equal  (ax.  3,  2Q). 

Therefore  the  complements  of  equal  angles  are  equal. 

51.  Theorem  5.  The  supplements  of  equal  angles 
are  equal.      [The  proof  is  left  to  the  student.  ] 


LINE- SEGMENTS  AND  ANGLES  23 

52.    Theorem  6.     //  two  adjacent  angles  are  supple- 
mental, then  their  exterior  sides  are  in  a  straight  line. 

Let  the  adjacent  angles  AOB  and  bog  be  supplemental. 


To  prove  that  OC  is  the  prolongation  of  OA. 

Suppose,  if  possible,  that  OC  is  not  the  prolongation  of  OA ; 
and  let  OC^  be  that  prolongation.  ^ 

Then  the  angle  BOC'  is  the  supplement  of  AOB  (49). 

Therefore  BOC^  and  BOC  are  equal,  being  supplements  of 
the  same  angle  (51). 

Hence  a  part  of  the  angle  BOC  \^  equal  to  the  whole  angle, 
which  is  impossible. 

Thus  the  supposition  made  is  proved  false,  since  it  leads,  by 
correct  reasoning,  to  an  absurdity. 

Therefore  OC'  and  OA  are  not  in  the  same  straight  line. 

In  the  same  way  it  can  be  shown  that  no  other  line  than 
OC  is  in  a  straight  line  with  OA. 
Therefore  OC  and  OA  are  in  one  straight  line. 

63.    Indirect   proof,    or  proof  by   exclusion.       It   may    be 

noticed  that  the  two  statements  above,  ^'OC  is  the  prolonga- 
tion of  0^,"  and  "  OC  is  not  the  prolongation  of  OA,"  are 
opposite  statements ;  i.e.  if  either  is  false,  the  other  is  true. 
Instead  of  proving  the  truth  of  the  first  statement  directly, 
it  was  easier  to  prove  the  falsity  of  its  opposite,  or,  in  other 
words,  "  to  exclude  the  opposite.'' 

The  process  of  proving  the  truth  of  a  statement  indirectly, 
by  proving  the  falsity  of  its  opposite,  is  called  indirect 
proof  or  proof  hy  exclusion. 

MCM.    ELEM.    GEOM. 3 


24  PLANE  GEOMETRY — BOOK  I 

54.  Converse  theorems.  There  is  a  close  relation  between 
52  and  49,  which  will  be  seen  more  clearly  if  the  latter  is 
stated  in  the  following  form ; 

If  two  concave  adjacent  angles  have  their  exterior 
sides  in  a  straight  line,  then  the  two  adjacent  angles 
are  supplemental. 

Art.  52  differs  from  this  by  having  the  hypothesis  and 
conclusion  interchanged.  The  interchange  of  hypothesis 
and  conclusion  is  called  conversion. 

Definition.     Two  theorems  are  said  to  be  converse  to  each 
other  when  the  hypothesis  of  each  is  the  conclusion  of  the 
other.     Two  converse  theorems  have  the  type-forms : 
If  A  is  Bj  then  C  is  Z) ;  If  C  is  D,  then  A  is  B. 

In  many  cases  two  converse  theorems  are  both  true ;  but 
there  are  cases  in  which  a  theorem  is  true  while  its  converse 
is  not  true.  The  truth  of  the  converse  is  not  a  logical  conse- 
quence of  the  truth  of  the  original  theorem,  but  always 
requires  separate  examination. 

55.  Definition.  Two  angles  that  are  situated  so  that  the 
sides  of  each  are  the  prolongations  of  the  sides  of  the  other 
are  said  to  be  vertically  opposite  angles. 

56.  Theorem  7.  Two  vertically  opposite  angles  are 
equal. 

Let  the  vertically  opposite  angles  AOB,  A' OB'  be  formed 
by  the  straight  lines  AOa'^  bob'. 

A'  b 


O 

B'  A. 

To  prove  that  the  angles  AOB  and  A' OB'  are  equal. 
The  angle  AOB  is  the  supplement  of  BOA'  (def.,  49) ;  and 
^OB'  is  also  the  supplement  of  BOA'. 

Now  supplements  of  the  same  angle  are  equal  (51). 
Therefore  the  angles  AOB  and  A' OB'  are  equal. 


TBIANGLES  25 

TRIANGLES 

57.  The  preceding  articles  treated  of  the  figure  formed  by 
two  intersecting  lines.  The  next  figure  in  order  of  simplicity 
is  that  formed  by  three  lines  each  of  which  intersects  the 
other  two. 

58.  Definitions.  A  plane  figure  formed  by  three  straight 
lines  that  inclose  a  portion  of  the  plane  surface  is  called  a 
triangle. 

These  lines  are  called  the  sides,  and  their  intersections 
the  vertices  of  the  triangle. 

Unless  otherwise  stated,  the  sides  will  be  taken  to  mean 
the  segments  lying  between  the  vertices  ;  but  they  may  also 
be  thought  of  as  indefinitely  prolonged. 

The  angles  formed  by  the  sides,  situated  toward  the 
interior  of  the  triangle,  are  called  the  interior  angles,  or 
simply  the  angles,  of  the  triangle. 

An  exterior  angle  of  the  triangle  is  the  concave  angle 
formed  by  one  of  the  sides  and  the  prolongation  of 
another. 

It  is  sometimes  convenient  to  regard  a  triangle  as  stand- 
ing on  a  selected  side.  We  then  call  that  side  the  base; 
the  two  angles  adjacent  to  it,  the  base  angles;  the  opposite 
angle,  the  vertical  angle;  the  sides  of  this  angle,  the  two 
sides  of  the  triangle ;  and  its  vertex,  the  vertex  of  the  tri- 
angle. 

An  isosceles  triangle  is  one  that  has  any  two  of  its  sides 
equal. 

The  vertex  common  to  the  equal  sides  of  an  isosceles 
triangle  is  called  the  vertex,  and  the  side  opposite  to  it  is 
called  the  base. 

An  equilateral  triangle  is  one  that  has  its  three  sides 
equal. 

A  scalene  triangle  is  one  that  has  no  two  of  its  sides 
equal. 


26  PLANE  GEOMETRY  —  BOOK  I 

Isosceles  Triangles 

59.  Theorem  8.  In  a  triangle,  if  two  sides  are  equal, 
then  the  angles  opposite  tJie  equal  sides  are  equal. 

Let  the  triangle  ABC  have  the  sides  AB  and  AC  equal. 

To  prove  that  the  angles  ABC  and  ACB  are  equal. 

Imagine  the  triangle  turned  over  on  itself 
so  that  AB  takes  the  position  AC,  and  AC 
takes  the  position  AB. 

Since  AB  and  AC  are  equal,  the  point  B 
takes  the  position  of  C,  and  C  takes  tlje  posi- 
tion of  B ;  hence  the  new  position  of  the  line 
BC  coincides  with  its  old  position  (Introd.  17). 

Therefore  the  angles  ABC  and  ACB,  being  superposable, 
are  equal. 

60.  Cor.  I.  If  the  equal  sides  AB  and  AC  are  extended 
through  B  and  C,  the  angles  below  the  base  are  equal. 

61.  Cor.  2.  If  two  angles  of  a  triangle  are  not  equal,  then 
the  opposite  sides  are  not  equal. 

Proof.    Suppose,  if  possible,  that  the  opposite  sides  are  equal. 
Then  the  angles  opposite  these  sides  are  equal  (59). 
This  is  inconsistent  with  the  hypothesis.    Therefore  the  supposi- 
tion made  is  false.    Hence  the  opposite  sides  are  not  equal. 

Converse  of  theorem  8, 

62.  Theorem  9.  In  a  triangle,  if  two  angles  are 
equal,  then  the  sides  opposite  the  equal  angles  are 
equal. 

Let  the  triangle  ABC  have  the  angles  B  and  C  equal. 
To  prove  that  the  sides  AB  and  AC  are  equal. 
Imagine  the  figure  turned  over  on  itself  so  that  B  falls  on 
C,  and  C  on  B. 


TRIANGLES  27 

Then  the  line  BA  takes  the  position  CA,  since  the  angles 
B  and  C  are  equal. 

Similarly  the  line  CA  takes  the  position  CB. 

Therefore  the  point  A,  being  the  intersection  of  BA  and 
CA,  falls  on  its  former  position. 

Hence  the  sides  BA  and  CA,  being  superposable,  are  equal. 

63.  Cor.  If  two  sides  of  a  triangle  are  not  equal,  then  the 
opposite  angles  are  not  equal.     [Prove  by  exclusion  as  in  61.] 

Equality  of  Triangles  —  Three  Primary  Cases 
Two  sides  and  included  angle, 

64.  Theorem  10.  //  two  triangles  have  two  sides 
and  the  included  angle  of  one  respectively  equal  to 
two  sides  and  the  included  angle  of  the  other,  the 
triangles  are  equal. 

Let  the  triangles  ABC  and  a'b^C^  have  the  sides  AB,  AC, 
and  the  angle  BAC,  respectively  equal  to  the  sides  a'b',  A'c', 
and  the  angle  b'a'c'. 

.A  A' 


To  prove  that  the  triangles  ABC  and  A^B^c'  are  equal. 

Imagine  the  angle  A  placed  on  the  equal  angle  A ',  the  side 
AB  taking  the  position  A'b',  and  AC  the  position  A'c'. 

Since  AB  equals  A'b',  and  AC  equals  A'c',  the  point  B  falls 
on  B',  and  C  on  c'.  Hence  the  line  BC  coincides  throughout 
with  B'C'  (Introd.  17). 

Therefore  the  triangles,  being  superposable,  are  equal. 

Note.  Sometimes  two  equal  triangles  cannot  be  superposed  with- 
out turning  one  of  them  over.     (See  73,  76.) 


28 


PLANE  GEOMETRY  —  BOOK  I 


Two  angles  and  the  included  aide, 

65.  Theorem  11.  If  two  triangles  have  two  angles 
and  the  intervening  side  of  one  respectively  equal  to 
two  angles  and  the  intervening  side  of  the  other,  the 
triangles  are  equal. 

Let  the  triangles  ABC  and  a'b'c'  have  the  angles  B,  c,  and 
the  side  BC,  respectively  equal  to  the  angles  b',  c\  and  the 
side  B'&. 


B  c       B'  c' 

To  prove  that  the  triangles  ABC  and  a'b'c'  are  equal. 

Imagine  the  angle  B  placed  on  the  angle  B\  the  side  BC 
taking  the  position  B'c'j  and  BA  the  position  b'a'. 

Since  BC  equals  B'c',  the  point  C  falls  on  C';  and  since 
the  angles  C  and  C'  are  equal,  the  line  CA  falls  on  c'a'. 

Therefore  the  points,  the  intersection  of  BA  and  CA,  falls 
on  A',  the  intersection  oi  B'A'  and  c'A'. 

Hence  the  triangles  are  superposable  and  equal. 

Three  sides, 

66.  Theorem  12.  If  two  triangles  have  three  sides 
of  one  equal  respectively  to  three  sides  of  tJie  other,  the 
triangles  are  equal. 

Let  the  triangles  ABC  and  A'b'c'  have  the  sides  AB,  BC, 
AC  equal  respectively  to  A' b',  b'c',  A'c'. 

B  B' 


TRIANGLES  29 

To  prove  that  the  triangles  ABC  and  A'b'c'  are  equal. 

Place  the  triangle  ABC  so  that  AC  coincides  with  its  equal 
A'C';  the  point  B  falling  at  the  same  side  as  B'. 
It  is  then  to  be  proved  that  B  falls  on  B'. 
This  is  evident  either  if  AB  falls  on  A'b',  or  BC  on  B'c'. 

Suppose  if  possible  that  neither  of  these  coincidences 
takes  place;  and  first  let  ABC  take  the  position  A'b"c', 
each  of  the  vertices  B'  and  B"  being  without  the  triangle 
to  which  the  other  belongs.     Join^'jB". 

Then  the  triangle  A'b'b"  is  isosceles  because  A'b'  equals 
A'B" ;  hence  the  angles  A'b'b"  and  A'b"b'  are  equal  (59). 

Therefore  the  angle  C'^'^"  is  less  than  the  angle  c'b"b' 
(the  former  being  less  than  one  of  the  equal  angles  and  the 
latter  being  greater  than  the  other). 

But  the  triangle  c'b'b"  is  isosceles,  since  the  sides  C'b' 
and  C'B"  are  equal ;  therefore  the  angles  C'b'b"  and  c'b"b' 
are  equal  (59). 

Hence  these  angles  are  at  the  same  time  unequal  and 
equal ;  which  is  impossible. 

Therefore  the  supposition  made  is  false. 

Next  let  one  of  the  vertices  B'  and  B"  lie  within  the  tri- 
angle to  which  the  other  belongs. 

It  may  be  proved  in  a  similar  manner  that  this  sup- 
position leads  to  an  impossibility.  [The  student  may  draw 
a  figure  and  prove.]     Therefore  the  point  B  falls  on  B'. 

Hence  the  triangles  ABC  and  a'b'c'  are  superposable,  and 
equal. 

Some  Fundamental  Constructions 

67.  A  geometrical  problem  is  a  proposition  whose  object 
is  the  construction  of  a  figure  which  shall  conform  to  certain 
prescribed  conditions. 

The  solution  consists:  (1)  in  showing  how  to  use  the 
ruler  and  compass  so  as  to  make  the  required  figure ;  (2)  in 
demonstrating  that  the  figure  so  constructed   satisfies  the 


30  PLANE  GEOMETRY  —  BOOK  I 

prescribed  conditions ;  and  (3)  in  discussing  what  limitations 
there  are  on  these  conditions  so  that  a  solution  may  be 
possible,  and  under  what  circumstances  there  may  be  more 
than  one  solution. 

68.  Problem  1.  On  a  given  finite  line  to  construct 
an  equilateral  triangle. 

Let  AB  be  the  given  line  on  which  it  is  required  to  con- 
struct an  equilateral  triangle. 

With  A  as  center  and  AB  as  radius, 
describe  the  arc  BLC  (post.  3,  Introd. 
29).  With  B  as  center  and  BA  as  radius, 
describe  the  arc  AMC.  Let  the  two 
arcs  intersect  in  C.     Draw  CA  and  CB. 

Then  ABC  is  an  equilateral  triangle. 

For  AC  and  AB  are  equal,  being  radii  of  the  same  circle ; 
and  BC  and  AB  are  equal,  being  radii  of  the  same  circle. 

Hence  the  triangle  ABC  has  its  three  sides  equal ;  and  it 
is  therefore  equilateral. 

Transference  of  line-segment, 

69.  Problem  2.  On  a  given  straight  line  to  lay  off  a 
part  equal  to  a  given  finite  straight  line. 

Let  LL'  be  the  given  line  from  which  it  is  required  to  lay 
off  a  part  equal  to  the  given  finite  line  AB. 


L P—Z J^ L' 

Take  any  point  0  on  the  line  LZ*';  and  with  O  as  center 
and  a  radius  equal  to  AB,  describe  a  circle  cutting  the  given 
line  in  the  points  P,  P'. 

Then  either  of  the  parts  OP,  OP'  answers  the  requirements 
of  the  problem,  since  they  are  each  equal  to  AB. 


TRIANGLES  31 

Bisection  of  line-segtnent, 

70.   Problem  3.     To  bisect  a  given  finite  straight  line ; 
that  is,  to  divide  it  into  two  equal  parts. 

Let  AA^  be  the  given  line  which  it  is  required  to  bisect. 


/ 

y1 

1 

^N 

w 

0   IB        ^ 

-^A 

N^ 

\ 

1          /' 

/   y 

r 

With  A  as  center,  and  any  convenient  radius  AB,  describe 
the  arc  GBC\  With  A^  as  center  and  an  equal  radius, 
describe  the  arc  CB^G\  Let  these  arcs  intersect  in  C,  C'. 
Draw  CC\  meeting  A  A'  at  O. 

Then  0  is  the  required  mid-point  of  AA\ 

For  in  the  triangles  AC&  and  A^CC\  the  sides  AC  and  A^C 
are  equal,  since  the  circles  have  equal  radii.  Similarly  the 
sides  AC'  and  A'c'  are  equal.  Also  the  side  CC'  is  common 
to  the  two  triangles. 

Therefore  the  angles  ACC'  and  A^CC'  are  equal  {^^). 

Again,  in  the  triangles  ACQ  and  A' CO,  the  sides  AC  and  A^C 
are  equal ;  the  side  CO  is  common ;  and  the  included  angles 
ACO  and  OCA'  have  been  proved  equal ;  therefore  AO  and  OA' 
are  equal  (64), 

Hence  AA'  is  bisected  at  0. 

Note.  Observe  that  if  the  radius  AB  is  taken  too  small  the  circles 
will  not  intersect.  It  is  always  possible  to  take  such  a  radius  that  the 
circles  shall  intersect  at  each  side  of  the  given  line  ;  this  may  be  in- 
ferred from  the  statements  with  regard  to  closed  figures  (Introd.  27). 
Experience  will  show  what  length  of  radius  is  preferable  for  conven- 
ience and  accuracy. 

71.   Cor.  I.     Prove  that  a  line-segment  has  only  one  mid-point 


32  PLANE  GEOMETRY  —  BOOK  I 

72.  Cor.  2.  If  three  finite  lines  are  such  that  the  difference 
of  the  first  and  second  is  equal  to  the  difference  of  the-  second 
and  third,  then  the  sum  of  the  first  ai\d  third  is  equal  to  double 
the  second. 


I:: h 


O  A        'B        C  D 

Outline.  On  an  indefinite  line  lay  off  OA,  OB,  OC,  equal  respec- 
tively to  the  given  lines.  Then  by  liypothesis  AB  and  BC  are  equal. 
Take  CD  equal  to  OA.     Prove  OD  equal  to  double  OB. 

Bisection  of  angle* 

73.  Problem  4.  To  bisect  a  given  angle;  that  is,  to 
divide  it  into  two  equal  parts. 

Let  AOB  be  the  given  angle  which  it  is  required  to  bisect. 


Lay  off  any  convenient  equal  segments  OM  and  ON. 
With  if,  N  as  centers  and  any  convenient   equal  radii, 
describe  arcs  intersecting  at  C.     Draw  OC. 

The  straight  line  OC  bisects  the  given  angle  AOB. 

To  prove  this,  draw  MC  and  NC. 

The  triangles  OMC  and  ONC  have  their  sides  respectively 
equal ;  therefore  the  angles  MOC  and  NOC  are  equal  (66). 

Ex.  1.     A  given  angle  has  only  one  bisector. 

Ex.  2.     Bisect  a  given  convex  angle. 

Ex.  3.     Bisect  a  given  straight  angle. 

Ex.  4.     Show  how  to  divide  a  given  angle  into  four,  eight,  sixteen, 
.  .  .  equal  parts. 


TRIANGLES 


33 


Erecting  perpendicular, 

74.  Problems.     To  erect  a  perpendicular  to  a  given 
line  at  a  given  point  of  the  line. 

[Bisect  the  straight  angle  by  the  method  of  Problem  4.] 

Dropping  perpendicular. 

75.  Problem  6.     To  drop  a  perpendicular  to  a  given 
line  from  a  given  point  not  on  the  line. 

Let  iZ/'  be  the  given  line,  and  0  the  given  point  from 
which  a  perpendicular  is  to  be  drawn. 


With  O  as  center  and  any  convenient  radius,  describe  an 
arc  cutting  the  given  line  in  the  two  points  M,  N. 

Bisect  MN  at  the  point  P  (70).     Draw  OP. 

Then  OF  is  the  required  perpendicular. 

For  the  triangles  MOP  and  NOP  have  their  sides  respec- 
tively equal  by  construction. 

Therefore  the  angles  MPO  and  NPO  are  equal  (pQ). 

Hence  these  angles  are  right  angles  (17). 

Therefore  OP  is  perpendicular  to  MN. 

76.  Discussion.  There  is  only  one  solution  to  this  problem. 
In  other  words : 

There  is  only  one  perpendicular 
from  a  given  point  to  a  given  line. 

Outline  proof.  Suppose,  if  possible,  that 
O^  is  a  second  perpendicular.  Prolong  OP^ 
making  PiY  equal  to  OP.  JoiniV^.  Prove 
that  the  angles  OQP  and  NQP  are  equal ; 
and  that  0  QN  is  a  straight  line  (52) .  Show 
the  absurdity  from  Art.  17  of  Introduction. 
Draw  conclusion.  N 


\ 


84 


PLANE  GEOMETRY  —  BOOK  I 


Transference  of  an  angle, 

77.  Problem  7.  At  a  given  point  in  a  given  straight 
line  to  construct  an  angle  equal  to  a  given  angle. 

Let  0  be  the  given  point  in  the  given  line  OA,  and  a'o'b' 
the  given  angle.  It  is  required  to  draw  a  line  OB  making 
the  angle  AOB  equal  to  a'o'b'. 


O  A        O'  A 

With  0  and  0'  as  centers,  and  any  convenient  equal  radii 

OA  and  0U\  draw  the  arcs  AB  and  A'b'.     Let  the  latter 

arc  cut  the  sides  of  the  given  angle  at  the  points  A\  B'. 

Draw  the  line  A'b'. 
With  A  as  center  and  a  radius  equal  to  A'b',  draw  an  arc 

cutting  the  arc  AB  at  the  point  B.     Draw  the  line  OB. 

This  line  OB  makes  with  OA  an  angle  equal  to  the  angle 
A'O'B'.     (Prove  by  66.) 

Note.     By  this  construction  the  isosceles  triangle  O'A'B'  is  trans- 
ferred to  the  position  OAB.    The  possibility  of  such  transference 

iatereect  ''  '''*^'  ^"""'^  ^'"""^  ^'  ^^  ^*  ""^^'^  ^^^  ^^^  ^^^« 

Transference  of  a  triangle, 

I'^'^rr  f  r  "'^'  ^^""^  ^'^  -^^^^  ^^  ^  ^^'^^^  ^^^^«^  ^^-^^ 

^A  B  )  iby  actual  construction  with  ruler  and  compassl 


0' 


TRIANGLES  35 

Inequalities  relating  to  Triangles 

The  following  six  theorems  with  their  corollaries  treat  of 
the  fundamental  inequalities  involving  the  angles  or  sides  of 
a  triangle. 

Exterior  and  interior  angles, 

79.  Theorem  13.  If  any  side  of  a  triangle  is  pro- 
longed, the  exterior  angle  is  greater  than  either  of 
the  two  opposite  interior  angles. 

Let  the  side  AB  of  the  triangle  ABC  be  prolonged  to  D. 


To  prove  that  the  exterior  angle  CBD  is  greater  than  either 

of  the  interior  angles  5^C  and  BCA. 

Bisect  BC  at  E.  Draw  AE  and  prolong  it  to  F,  making  EF 
equal  to  AE.     Join  BF. 

In  the  triangles  AEC  and  BEF,  the  sides  AE  and  EC  are 
respectively  equal  to  EF  and  EB  (const.),  and  the  included 
angles  AEG  and  BEF  are  equal  (p^). 

Therefore  the  angles  ACE  and  FBE  are  equal  (64). 

Now  the  angle  CBD  is  greater  than  its  part  FBE. 

Therefore  the  angle  CBD  is  greater  than  ACE  (Introd.  35). 

In  the  same  way  (by  bisecting  the  side  AB  and  making  a 
similar  construction)  the  angle  ABG  can  be  proved  greater 
than  BAC]  hence  the  vertically  opposite  angle  CBD  is  also 
greater  than  BAC. 

Therefore  the  exterior  angle  CBD  is  greater  than  either  of 
the  interior  and  opposite  angles. 


36  PLANE  GEOMETRY  —  BOOK  I 

80.  Cor.  The  mm  of  two  angles  of  a  triangle  is  less  than  a 
straight  angle. 

Outline.  Prove  the  sum  of  CAB  and  CBA  less  than  the  sum  of 
CBD  and  CBA. 

Ex.  If  a  triangle  has  one  right  angle  or  obtuse  angle,  its  remaining 
angles  are  acute. 

Definitions.  A  triangle  is  called  a  right  triangle  when 
one  of  its  angles  is  a  right  angle  ;  an  obtuse  triangle  when 
one  angle  is  obtuse ;  an  acute  triangle  when  all  its  angles 
are  acute. 

Ex.    Prove  that  an  equilateral  triangle  is  an  acute  triangle. 

Unequal  sides  and  unequal  angles, 

81.  Theorem  14.  If  one  side  of  a  triangle  is  greater 
than  another,  the  angle  opposite  the  greater  side  is 
greater  than  the  angle  opposite  the  less. 

In  the  triangle  ABC,  let  the  side  ^c  be  greater  than  AB. 


G ■ ^^ 

To  prove  that  the  angle  ABC  is  greater  than  ACB. 

From  the  greater  side  lay  off  a  part  AD  equal  to  the  less 
side  AB.     Draw  db. 

In  the  isosceles  triangle  abd,  the  angles  abb  and  ABB 
are  equal  (59). 

Now  the  angle  ADB  is  greater  than  the  interior  angle  ACB 
(79). 

Therefore  ABD  is  greater  than  ACB  (30,  ax.  8). 
Hence  the  whole  angle  abc  is  greater  than  ACB. 


TRIANGLES  37 

82.  Combined  statement.  For  brevity,  denote  the  angles 
of  a  triangle  by  the  single  capital  letters  A,  B,  c,  and  the  op- 
posite sides  by  the  corresponding  small  letters  a,  b,  c ;  then, 
by  Theorems  8  and  14  the  following  statements  are  true: 

(1)  If  b  is  equal  to  c,  then  B  is  equal  to  C; 

(2)  If  b  is  greater  than  c,  then  B  is  greater  than  C; 

(3)  If  b  is  less  than  c,  then  B  is  less  than  a 

These  three  statements  may  be  conveniently  combined  into 
one  general  statement  as  follows : 

According  as  one  side  of  a  triangle  is  greater  than, 
equal  to,  or  less  than,  another  side,  so  is  tJie  angle 
opposite  the  first  side  greater  than,  equal  to,  or  less 
than,  the  angle  opposite  the  second  side. 

Here  the  word  If  is  replaced  by  the  distributive  phrase 
According  as,  and  the  word  then  by  the  words  so  is. 

Converse  of  81, 

83.  Theorem  15.  If  one  angle  of  a  triangle  is 
greater  than  another,  then  the  side  opposite  the 
greater  angle  is  greater  than  the  side  opposite  the  less. 

In  the  triangle  ABC,  let  the  angle  ABC  be  greater  than  ^  OB. 

A 


C B 

To  prove  that  the  side  ^C  is  greater  than  AB. 

The  side  AC  is  either  equal  to,  less  than,  or  greater  than, 
the  side  AB. 

Now^C  is  not  equal  to  AB-,  for  then  the  angle  B  would  be 
equal  to  C  (59),  contrary  to  the  hypothesis. 

Again,  ^C  is  not  less  than  AB ;  for  then  the  angle  B  would 
be  less  than  C  (81),  contrary  to  the  hypothesis. 

It  only  remains  that  the  side  AC  is  greater  than  AB. 


38  PLANE  GEOMETRY  —  BOOK  I 

84.  Combined  statement.  By  62  and  83  the  following 
statements  are  true: 

If  B  is  equal  to  C,  then  b  is  equal  to  c; 
If  B  is  greater  than  C,  then  b  is  greater  than  c ; 
■   If  /?  is  less  than  C,  then  b  is  less  than  c. 

These  three  statements  are  respectively  converse  to  those 
of  Art.  82 ;  and,  like  them,  may  be  combined  into  one  com- 
plete statement  as  follows : 

According  as  one  angle  of  a  triangle  is  greater  than, 
equal  to,  or  less  than,  another,  so  is  the  side  opposite 
the  first  angle  greater  than,  equal  to,  or  less  than,  the 
side  opposite  the  second  angle. 

Perpendicular  and  oblique  lines, 

85.  Theorem  16.  Of  all  the  straight  lines  that  can 
he  draum  from  a  given  point  to  a  given  line  : 

(1)  t?ie  perpendicular  is  the  least; 

(2)  any  two  that  make  equal  angles  with  the  per- 
pendicular are  equal; 

(3)  one  that  makes  a  greater  angle  with  the  perpen- 
dicular is  greater  than  one  that  makes  a  less  angle. 

Let  OP  be  the  perpendicular 
from  the  given  point  0  to  the 
given  line  LL\  Let  ON,  OQ  be 
any  lines  making  equal  angles 
NOP,  QOP,  with  OP.  Let  OR 
make  with  OP  the  angle  POR 
greater  than  the  angle  POQ  or    L  N     P     Q      B        i^ 

NOP. 

(1)   To  prove  that  OP  is  less  than  OQ. 

The  angle  OQP  is  less  than  the  exterior  angle  OPL  (79). 
!N'ow  the  angles  OPL  and  OPQ  are  equal,  being  right  angles. 
Hence  OQP  is  less  than  OPQ;  therefore  the  opposite  side 
OP  is  less  than  OQ  (83). 


TRIANGLES  39 

(2)  To  prove  that  the  lines  ON,  OQ  are  equal.    [Apply  65.] 

(3)  To  prove  that  OR  is  greater  than  OQ  or  ON 

The  angle  OQR  is  greater  than  the  right  angle  OPR  (79) ; 
which  equals  the  right  angle  OPN\  which  is  greater  than 
the  interior  angle  ORQ  (79). 

Hence  the  angle  OQR  is  greater  than  ORQ  (30)  ;  and  there- 
fore the  side  OR  is  greater  than  OQ  (83). 

86.  Cor.  I7i  an  isosceles  triangle,  a  line  joining  the  vertex 
to  any  point  in  the  base  is  less  than  either  side  ;  and  a  line 
joining  the  vertex  to  any  point  in  the  base  extended  is  greater 
than  either  side. 

Sum  of  two  sides, 

87.  Theorem  17.  Any  side  of  a  triangle  is  less  than 
the  sum  of  the  other  two. 

Let  ABC  be  a  triangle. 

To  prove  that  any  side  AR  is  less  than  y1 

the  sum  of  the  other  two  sides  AC  and 
BC. 

Prolong  the  side  AC  until  the  pro- 
longation CD  equals  the  side  CB;  and 
draw  BD. 

In  the  isosceles  triangle  BCD,  the  angle  CBD  equals  CDB. 

Hence  the  whole  angle  ABD  is  greater  than  the  angle  CDB. 

Therefore,  in  the  triangle  ADB,  the  side  AD,  opposite  the 
greater  angle,  is  greater  than  the  side  AB  (83). 

Now  AD  equals  the  sum  of  ^C  and  CD,  which  equals  the 
sum  of  ^C  and  CB. 

Therefore  the  sum  of  ^C  and  CB  is  greater  than  AB. 

88.  Cor.  I.  Any  side  of  a  triangle  is  greater  than  the 
difference  of  the  other  two. 

89.  Cor.  2.  Any  straight  line  is  less  than  the  sum  of  the 
parts  of  a  broken  line  having  the  same  extremities. 

MCM.  ELEM.   GEOM. — 4 


40  PLANE  GEOMETRY  —  BOOK  I 

90.  Cor.  3.  If  from  the  ends  of  a  side  of  a  tnangle  two 
straight  lines  are  drawn  to  a  point  within  the  triangle,  their 
sum  is  less  than  the  sum  of  the  other  two  sides  of  the  tnangle. 

Definition.  The  sum  of  the  three  sides  is  called  the  perim- 
eter of  the  triangle. 

Ex.  The  sum  of  the  lines  joining  any  point  within  a  triangle  to  the 
three  vertices  is  less  than  the  perimeter  of  the  triangle,  and  greater 
than  half  the  perimeter. 

A  case  of  unequal  triangles, 

91.  Theorem  18.  If  two  triangles  have  two  sides 
of  one  respectively  equal  to  two  sides  of  the  other,  and 
the  included  angle  in  the  first  greater  than  the  in- 
cluded angle  in  the  second,  then  the  third  side  of  the 
first  is  greater  than  the  third  side  of  the  second. 

Let  the  two  triangles  ABC  and  A^B'c'  have  the  sides  AB,  BC 
respectively  equal  to  A^B\  B'&y  and  the  included  angle  ABC 
greater  than  a'b'C'. 

B 


To  prove  that  the  third  side  ^C  is  greater  than  A'C'. 

Draw  the  line  BC"  making  the  angle  ABC"  equal  to  the 
less  angle  B'.     Take  BC"  equal  to  B'c* ;  and  draw  AC",  CC". 

First  let  the  point  C"  fall  within  the  triangle  ABC.  Let 
M  and  N  be  points  on  the  prolongations  of  BC  and  BC". 

The  triangles  ABC"  and  a'b'c'  are  equal ;  and  the  sides 
AC"y  A'C'  are  equal  (64). 

In  Ijhe  isosceles  triangle  BCC",  the  angles  C&'N'  and  C"CM 
below  the  base  are  equal  (60). 


TRIANGLES 


41 


Hence  the  whole  angle  AC"C,  being  greater  than  one  of 
the  equal  angles,  is  greater  than  ACC",  which  is  a  part  of  the 
other.  Therefore,  in  the  triangle  AC"C,  the  side  AC,  being 
opposite  the  greater  angle,  is  greater  than  AC"  (83). 

Therefore  AC  is  greater  than  A'c'. 

Next  let  C"  fall  without  the  triangle  ABC. 


The  proof  for  the  second  figure  is  left  to  the  student ;  also  the  con- 
sideration of  the  intermediate  case  in  which  C"  falls  on  AC. 

Combined  statement, 

92.  Cor.  I.  If  two  triangles  have  two  sides  of  one  equal 
to  two  sides  of  the  other,  then  according  as  the  vertical  angle 
of  the  first  is  greater  than,  equal  to,  or  less  than,  the  vertical 
angle  of  the  second,  so  is  the  base  of  the  first  greater  than, 
equal  to,  or  less  than,  the  base  of  the  second.  [Combine  64 
and  91.] 

93.  Cor.  2  (Converse  of  91).  If  two  triangles  have  two 
sides  of  the  first  equal  respectively  to  two  sides  of  the  second, 
and  the  base  of  the  first  greater  than  the  base  of  the  second, 
then  the  vertical  angle  of  the  first  is  greater  than  that  of  the 
second.     [Prove  by  exclusion,  using  92 ;  see  83.] 

Coinhined  statement, 

94.  Cor.  3.  If  two  triangles  have  two  sides  of  the  first 
equal  to  two  sides  of  the  second,  then,  according  as  the  base 
of  the  first  is  greater  than,  equal  to,  or  less  than,  the  base  of 
the  second,  so  is  the  vertical  angle  of  the  first  greater  than, 
equal  to,  or  less  than,  the  vertical  angle  of  the  second.    [66,  93.] 


42  PLANE  GEOMETRY — BOOK  I 

Equality  of  Triangles.  —  Two  Secondary  Cases 

Two  angles  and  the  aide  opposite  one, 

95.  Theorem  19.  If  two  triangles  have  two  angles 
of  one  respectively  equal  to  two  angles  of  the  other,  and 
the  sides  opposite  one  pair  of  angles  equal  in  each  tri- 
angle, then  the  triangles  are  equal. 

Let  the  two  triangles  ABC  and  A^B'cf  have  the  angles  5,  C, 
and  the  side  AB  respectively  equal  to  the  angles  B\  cf,  and 
the  side  a'b'.  a  a' 


B  C"  c  B'  C 

To  prove  that  the  triangles  are  equal. 

Place  the  triangle  A^B'c^  on  ABC  so  that  A^B^  falls  on  its 
equal  AB,  and  the  angle  A'b'C^  on  its  equal  ABC\  then  the 
line  B'C'  falls  in  the  line  BC. 

Suppose,  if  possible,  that  B'c'  is  not  equal  to  BC,  and  that 
the  point  C'  falls  on  c"  instead  of  falling  on  C. 

Then  the  triangles  a'b'c^  and  ABC'^  have  the  sides  ^'5', 
B'C^,  and  the  included  angle  a'b'c',  respectively  equal  to  AB, 
BC",  ABC".  Therefore  these  triangles  are  equal;  and  the 
angles  AC"B  Sind  A'c'b'  are  equal  (64). 

But  the  angles  a'c'b'  and  ACB  are  equal  by  hypothesis. 

Therefore  the  angles  AC"B  and  ACB  are  equal;  which  is 
impossible  (79). 

Similar  reasoning  applies  if  C"  falls  at  the  other  side  of  C. 

Thus  the  supposition  that  C'  does  not  fall  on  c  is  false  j 
hence  C'  falls  on  C,  and  the  triangle  A'b'c'  on  ABC. 

Therefore  the  triangles  ABC  and  a'b'c'  are  equal. 

96.  Cor.  The  tivo  perpendiculars  drawn  from  any  point  in 
the  bisector  of  an  angle  to  the  sides  of  the  angle  are  equal. 


TRIANGLES 


43 


Two  sides  and  the  angle  opposite  one. 


97.  Theorem  20.  //  two  triangles  have  two  sides 
of  one  respectively  equal  to  two  sides  of  the  other,  and 
the  angle  opposite  one  of  these  equal  to  the  correspond- 
ing angle  in  the  other  triangle,  then  the  angles  opposite 
to  the  other  pair  of  equal  sides  are  equal  or  supple- 
mental ;  and  if  equal,  the  triangles  are  equal. 

Let  the  two  triangles  ABC  and  A'b'c'  have  the  sides  AB 
and  BC  respectively  equal  to  the  sides  A'B^  and  ^'c/,  and 
the  angle  BAC  equal  to  the  angle  b'a'C'. 


To  prove  that  the  angles  BCA  and  B^c'A'  are  either  equal 
or  supplemental. 

The  tvs^o  sides,  AC  and  A'c',  are  either  equal  or  unequal. 

If  they  are  equal  (as  in  fig.  1),  the  triangles  are  equal  in 
all  their  parts  (66). 

If  they  are  unequal  (as  in  fig.  2),  let  ^C  be  the  greater. 
Lay  off  AC"  equal  to  A'c';  and  draw  BC". 

In  the  triangle  ABC"  and  A'b'c',  the  sides  AB  and  AC"  and 
the  included  angle  BAC"  are  equal  respectively  to  the  sides 
a'b'  and  A'c'  and  the  included  angle  B'a'c'.  Therefore  the 
angles  BC"A  and  B'c'A'  are  equal,  and  the  sides  BC"  and  B'C* 
are  equal  (64). 

Therefore  BC"  equals  BC-,  and  hence  the  angles  BCC"  and 
BC"C  are  equal  (59). 

Now  BC"C  is  the  supplement  of  BC"A;  therefore  BCA,  be- 
ing equal  to  BC"C,  is  equal  to  the  supplement  of  BC"A,  and 
hence  equal  to  the  supplement  of  b'c'a'. 


44  PLANE  GEOMETRY  —  BOOK  I 

98.  Cor.  I.  If  two  triangles  have  tivo  sides  of  one  respec- 
tively equal  to  two  sides  of  the  other,  and  the  angle  opposite 
one  of  these  sides  equal  to  the  corresponding  angle  in  the  other 
triangle,  then  the  triangles  are  equal  : 

(1)  If  the  two  angles  given  equal  are  right  angles  or  obtuse 
angles  ; 

(2)  if  the  angles  opposite  to  the  other  two  equal  sides  are 
both  acute,  or  both  obtuse,  or  if  one  of  tJiem  is  a  right  angle  ; 

(3)  if  the  side  opposite  the  given  angle  in  each  triangle  is  not 
less  than  the  other  given  side. 

99.  Cor.  2.  Jlie  perperidicular  from  the  vertex  of  an  isos- 
celes triangle  to  the  base  bisects  both  the  base  and  the  vertical 
angle.  Conversely,  the  perpendicular  bisector  of  the  base  of  an 
isosceles  triangle  passes  through  the  vertex. 

100.  Cor.  3.  If  the  two  peipendiculars  drawn  from  a  point 
to  the  sides  of  an  angle  are  equal,  then  the  point  is  on  the  bi- 
sector of  the  angle.     (Converse  of  96.) 

EXERCISES 

1.  Summarize  the  five  cases  of  the  equality  of  two  triangles. 

2.  The  bisectors  of  two  adjacent  supplemental  angles  are  perpen- 
dicular to  each  other. 

3.  The  bisectors  of  two  adjacent  conjunct  angles  are  in  the  same 
straight  line. 

4.  The  bisectors  of  two  vertically  opposite  angles  are  in  the  same 
straight  line. 

5.  The  lines  drawn  from  the  extremities  of  the  base  of  an  isosceles 
triangle  to  the  middle  points  of  the  opposite  sides  are  equal. 

6.  The  bisector  of  the  vertical  angle  of  an  isosceles  triangle  bisects 
the  base  at  right  angles. 

7.  If  the  bisector  of  an  angle  of  a  triangle  is  perpendicular  to  the 
opposite  side,  the  triangle  is  isosceles. 

8.  If  the  perpendicular  from  a  vertex  to  the  opposite  side  bisects 
that  side,  then  the  triangle  is  isosceles. 

9.  If  two  triangles  have  a  common  base,  and  if  the  vertex  of  the 
second  triangle  is  within  the  first,  or  on  a  side  of  the  first,  then  the 
vertical  angle  of  the  first  triangle  is  less  than  that  of  the  second. 

10.   Three  equal  lines  cannot  be  drawn  from  a  point  to  a  line. 


SUMMARY  OF  TYPES   OF  INFERENCE  45 

SUMMARY  OF  TYPES  OF  INFERENCE 

The  foregoing  propositions  have  illustrated  various 
methods  of  drawing  conclusions  from  given  premises.  It  is 
now  time  for  us  to  consider  some  of  the  essential  features  of 
the  modes  of  inference  we  have  been  using,  and  to  see  the 
simple  logical  principles  that  underlie  them.  It  will  be  seen 
that  there  are  a  few  general  type-forms  which  appear  again 
and  again  under  various  modes  of  expression;  and  the 
student  will  thus  early  learn  to  recognize  the  logical  equiv- 
alence of  certain  statements  that  differ  only  in  form ;  also 
to  distinguish  between  different  statements  that  may  seem 
to  be  alike;  and  gradually  will  come  to  see  the  legitimate 
conclusions  that  can  be  inferred  from  any  given  premises. 

101.  Related  statements.  Hereafter  when  the  word  "  state- 
ment'' is  used  without  qualification,  it  will  be  understood 
to  mean  a  simple  assertion  of  the  form  "^  is  -B." 

It  has  been  seen  that  a  "  theorem  "  is  made  up  of  two  such 
statements  placed  in  a  certain  relation  to  each  other,  the 
relation  of  hypothesis  (or  antecedent)  to  conclusion  (or  con- 
sequent). 

When  we  say  that  the  theorem  is  "  true,"  we  do  not  mean 
that  either  statement  is  true  in  itself,  but  only  that  the  con- 
sequent is  true  whenever  the  antecedent  is  true. 

This  may  be  conveniently  expressed  by  saying  that  the 
truth  of  the  consequent  is  a  necessary  result  of  the  truth  of 
the  antecedent. 

When  any  two  statements  are  related  to  each  other  so  that 
the  truth  of  each  is  a  necessary  consequence  of  the  truth  of 
the  other,  they  are  said  to  be  equivalent  statements. 

For  instance,  the  two  statements, 
X  is  equal  to  r, 
half  X  is  equal  to  half  r, 
are  equivalent.     When  either  is  true,  so  is  the  other ;  and 
hence  when  either  is  false,  so  is  the  other. 


46  PLANE  GEOMETRY  —  BOOK  I 

Two  statements  are  said  to  be  partially  equivalent  when  the  truth  of 
one  is  a  necessary  consequence  of  the  truth  of  the  other,  but  the  truth 
of  the  latter  not  a  necessary  consequence  of  the  truth  of  the  former. 

For  instance,  the  two  statements, 

X  is  greater  than  F, 

X  is  greater  than  half  T, 

are  partially  equivalent.  The  truth  of  the  second  is  a  necessary  conse- 
quence of  the  truth  of  the  first,  but  the  truth  of  the  first  is  not  a  necessary 
consequence  of  the  truth  of  the  second.  When  the  second  is  true, 
the  first  may  be  trae  or  false. 

Two  statements  are  said  to  be  independent  when  neither 
is  a  necessary  consequence  of  the  other. 
For  instance,  the  two  statements, 

X  is  greater  than  F, 

X  is  less  than  double  F, 

are  independent.  When  either  is  true,  the  other  may  be 
true  or  false. 

Two  statements  are  said  to  be  inconsistent  when  they 
cannot  both  be  true  at  the  same  time. 

Inconsistency  is  of  two  kinds,  opposition  and  partial 
opposition. 

Two  statements  are  said  to  be  opposite,  if,  when  either 
is  true  the  other  is  false,  and  when  either  is  false  the  other 
is  true. 

For  instance,  the  two  statements, 

X  is  equal  to  F, 

X  is  not  equal  to  F, 
are  opposite. 

Two  statements  are  said  to  be  partially  opposite,  if  when  either  is 
true  the  other  is  false,  and  when  either  is  false  the  other  may  be  true 
or  false. 

For  instance,  the  two  statements, 

X  is  equal  to  F, 
X  is  less  than  F, 


SUMMARY  OF  TYPES  OF  INFERENCE  47 

are  inconsistent  but  only  partially  opposite.  They  cannot  be  true  at 
the  same  time,  but  they  may  be  false  at  the  same  time,  for  there  is  a 
third  alternative,  X  may  be  greater  than  Y. 

In  the  preceding  case  there  is  no  third  alternative  to  the  two  state- 
ments, for  Xis  either  equal  to  I"  or  not  equal  to  Y. 

102.  Reductio  ad  absurdum.  It  was  remarked  in  53  that 
instead  of  proving  the  truth  of  a  statement  directly  it  is 
sometimes  easier  to  prove  the  falsity  of  its  opposite.  Ex- 
amples of  such  indirect  proof  have  occurred  in  52,  61,  66, 
83,  93,  95.  The  usual  way  of  conducting  an  indirect  proof 
is  to  suppose  "  for  the  sake  of  argument "  (as  in  theor.  6) 
that  the  opposite  of  the  desired  conclusion  is  true.  Combine 
this  provisional  supposition  with  the  given  hypothesis ;  and 
then  show  that  it  leads  by  correct  reasoning  to  a  conclusion 
which  is  inconsistent  with  something  previously  accepted  as 
true.  There  must  then  be  an  error  somewhere.  If  no  flaw 
exists  in  the  reasoning,  then  the  error  must  lie  in  the  pro- 
visional hypothesis.  The  provisional  supposition  is  then 
declared  to  be  false ;  and  its  opposite  (i.e.  the  desired  con- 
clusion) is  finally  pronounced  true. 

This  mode  of  reasoning  is  called  reductio  ad  dbsurdum, 
which  may  be  briefly  defined  as  the  process  of  proving 
the  truth  of  a  statement  by  reducing  its  opposite  to  an 
absurdity. 

103.  Conversion.  The  relation  of  a  theorem  to  its  con- 
verse was  explained  in  54.  Examples  of  pairs  of  converse 
theorems  have  presented  themselves  in  59  and  62,  81  and  83, 
91  and  93,  96  and  100.  In  no  case  was  the  truth  of  the 
converse  theorem  a  direct  logical  inference  from  the  truth 
of  the  original  theorem,  but  required  separate  investigation. 

This  inquiry  is  of  the  same  nature  as  seeking  the  cause  of  a  given 
effect  in  physical  science. 

In  a  theorem,  we  may  regard  the  hypothesis  as  the  cause,  and  the 
conclusion  as  the  effect. 

The  question  to  be  decided  is  this :  Can  we  from  the  presence  of  the 
effect  infer  the  presence  of  this  particular  cause  ? 


48  PLANE  GEOMETRY  —  BOOK  I 

The  doubt  arises  from  the  fact  that  the  same  effect  may  follow  from 
different  causes. 

In  order  to  prove  that  any  particular  fact  is  the  true  cause  it  is  suf- 
ficient to  show  that  if  this  fact  is  not  present  the  effect  in  question  does 
not  follow. 

Thus  from  the  fact  that  "  C  is  i>  "  we  can  infer  the  fact  that  "  ^  is 
5"  if  we  can  show  that 

"  when  A  is  not  B,  C  is  not  D." 

This  would  exclude  the  supposition  that  "J.  is  not  J5,"  leaving  the 
only  alternative  that  "  ^  is  5." 

The  method  of  exclusion  is  usually  the  most  convenient  method  of 
proving  the  converse  of  a  theorem.  This  is  exemplified  in  the  proofs 
of  52  and  83. 

Conversion  is  thus  not  a  process  of  logical  inference.  The  truth  or 
falsity  of  the  result  is  to  be  proved  by  a  fresh  appeal  to  geometric  facts. 

For  instance,  the  truth  of  83  was  derived  by  the  method 
of  exclusion  from  the  combined  statement  of  82.  The 
proof  consisted  in  showing  that  there  were  only  three 
alternatives,  namely,  the  three  hypotheses  of  Art.  82,  then 
excluding  two  of  these  alternatives  by  showing  that  each 
leads  to  a  contradiction  of  the  assumed  fact,  thus  leaving 
the  third  alternative  as  the  true  one. 

By  the  same  method  the  converse  of  each  of  the  state- 
ments of  Art.  82  can  be  proved  by  using  the  other  two. 

As  this  method  of  conversion  will  be  frequently  used,  it  is 
now  convenient  to  state  and  prove  the  general  principle 
involved,  once  for  all,  so  that  it  may  be  available  for 
reference. 

104.   Rule  of  conversion.     The  principle  may  be  stated  in 

general  terms  as  follows :  * 

"If  the  hypotheses  of  a  group  of  demonstrated  theorems 
be  exhaustive  —  that  is,  form  a  set  of  alternatives  of  which 
one  must  be  true;  and  if  the  conclusions  be  mutually 
exclusive  —  that  is,  be  such  that  no  two  of  them  can  be  true 

*  This  admirable  statement  is  quoted  from  the  Syllabus  of  the  Asso- 
ciation for  the  Improvement  of  Qeometrical  Teaching  (London). 


SUMMARY  OF  TYPES   OF  INFERENCE  49 

at  the  same  time,  then  the  converse  of  every  theorem  of  the 
group  will  necessarily  be  true." 

For  instance,  suppose  the  following  theorems  have  been 

demonstrated : 

(1)  When  A  is  B,    C  is  i); 

(2)  When  A  is  B',  (7  is  2)'; 

(3)  When  A  is  B",C  is  D" ; 

and  suppose  it  is  also  known  that  A  must  have  one  of  the 
qualities  B,  B',  B" ;  and  that  C  cannot  have  more  than  one 
of  the  qualities  D,  D',  D".  We  can  then  prove  that  the  fol- 
lowing three  converse  theorems  are  all  true : 

When  C  is  D,    A  is  B; 

When  C  is  D',  A  is  B'; 

When  C  is  Z)",  A  is  B", 

Take,  for  instance,  the  third  one. 

Suppose,  if  possible,  that 

When  C  is  B",  A  is  not  B". 

Then,  since  the  hypotheses  of  (1),  (2),  (3)  are  exhaustive, 
A  must  be  either  B  or  B'. 

Therefore,  by  (1)  and  (2),  C  must  be  either  D  ot  D'. 

Then,  since  the  conclusions  of  (1),  (2),  (3)  are  exclusive, 
C  cannot  he  D". 

Hence,  when  C  is  D",  C  is  not  B",  which  is  impossible. 

Therefore  the  supposition  made  is  false.     Hence, 
When  C  is  D",  A  is  B". 

Thus  the  "  Rule  of  Conversion  "  is  established. 

Case  of  two  alternatives.  In  82  and  84  there  are  three 
alternatives.  The  rule  of  conversion  can  likewise  be  applied 
to  the  case  of  tivo  demonstrated  theorems  whose  hypotheses 
are  exhaustive  and  whose  conclusions  are  exclusive. 

Ex.     Assuming  the  truth  of  the  following  two  theorems  : 
When  A  is  B,   C  is  D ; 
When  A  is  not  B,  C  is  not  D; 
apply  the  rule  of  conversion  to  prove  the  truth  of  the  converse  of  each. 


50  PLANE  GEOMETRY  —  BOOK  I 

105.  Equivalent  theorems.  If  two  theorems  are  such  that 
each  follows  logically  from  the  other,  the  two  theorems  are 
said  to  be  equivalent. 

If  one  of  two  equivalent  theorems  is  true,  so  is  the  other. 
Hence,  if  one  of  them  is  false,  so  is  the  other. 

Two  equivalent  theorems  are  not  in  strictness  diiferent  the- 
orems, but  diiferent  ways  of  enunciating  the  same  theorem. 

A  simple  example  of  such  equivalence  is  seen  on  comparing  59  and 
61.  It  will  appear  in  the  next  Article  that  each  of  these  theorems  is  a 
logical  consequence  of  the  other.  They  are  related  to  each  other  in 
a  peculiar  way.  The  first  asserts  that  a  certain  hypothesis  leads  to  a 
certain  conclusion  ;  the  second  asserts  that  the  opposite  of  that  con- 
clusion leads  to  the  opposite  of  that  hypothesis.  Theorems  related  in 
this  way  are  said  to  be  contraposed  to  each  other. 

106.  Contraposition.  Two  theorems  are  said  to  be  contra' 
posed  to  each  other  when  the  hypothesis  of  each  is  the 
opposite  of  the  conclusion  of  the  other.  They  may  be  rep- 
resented by  the  type-forms  : 

(1)  If  A  is  5,  then  Cis  D; 

(2)  If  C  is  not  Dj  then  A  is  not  B. 

Each  of  these  theorems  is  called  the  contraposite  of  the 
other. 

Equivalence  of  contraposed  theorems.  Each  of  two 
contraposed  theorems  follows  logically  from  the  other. 

For  instance,  assuming  the  first  to  be  true,  we  may  infer 
the  truth  of  the  second,  by  the  method  of  exclusion,  as 
follows : 

To  prove  that 

when  C  is  not  2),  then  A  is  not  B. 

Suppose,  if  possible,  that 

when  C  is  not  D,  then  A  is  B. 

Now,  when  A  is  B,  then  C  is  Z),  by  (1),  whose  truth  was 
assumed. 


SUMMARY  OF  TYPES  OF  INFERENCE  51 

Hence,  when  C  is  not  D,  then  C  is  jD  ;  which  is  impossible, 
since  these  two  statements  are  opposite. 

Therefore  the  supposition  made  is  false.     Hence, 

when  C  is  not  2),  then  A  is  not  B. 

In  the  same  way  the  first  theorem  may  be  shown  to  be  a 

logical  consequence  of  the  second- 
Two  contraposed  theorems  may  thus  be  regarded  as  different  ways 

of  expressing  the  logical  dependence  of   one  fact  or  property  upon 

another. 

The  first  form  asserts  that  the  property  expressed  by  the  statement 

"J.  is  jB"  is  always  accompanied  by  the  property  expressed  by  the 

statement  "  O  is  D." 

The  second  form  asserts  that  the  absence  of  the  latter  property 

shows  the  absence  of  the  former. 

Sometimes  one  form  may  be  more  convenient,  and  some- 
times the  other,  according  to  the  use  we  wish  to  make  of  the 
fact  in  question. 

Thus  contraposition  is  a  process  of  purely  logical  inference. 
It  requires  no  fresh  appeal  to  geometric  facts. 

107.  The  four  related  types.  Comparing  59,  its  converse  62, 
and  their  two  contraposites  (61,  63),  we  have  a  group  of  four  theorems 

of  the  types : 

(1)  If  A  is  B,  then  C  is  D  ; 

(2)  If  C  is  not  D,  then  A  is  not  B ; 

(3)  If  O  is  D,  then  ^  is  5; 

(4)  If  A  is  not  B,  then  C  is  not  D. 

Of  these  four  theorems 

the  first  and  second  are  contraposed, 
the  first  and  third  are  converse, 
the  third  and  fourth  are  contraposed, 
the  second  and  fourth  are  converse. 

We  have  now  to  introduce  a  third  term,  the  word  obverse,  to  ex- 
press the  relation  that  exists  between  the  fii-st  theorem  and  the  fourth, 
and  also  between  the  second  and  the  third. 


52  PLANE  GEOMETRY  —  BOOK  I 

108.  Obversion.  Two  theorems  are  said  to  be  obverse  to  each 
other  when  their  hypotheses  are  opposite,  and  their  conclusions  also 
opposite. 

To  ohvert  a  theorem  is  to  change  the  hypothesis  into  its  opposite, 
and  the  conclusion  into  its  opposite. 

The  obverse  of  a  theorem  is  then  the  contraposite  of  its  converse. 
For  instance,  the  converse  of  (1)  is  (3),  and  the  contraposite  of  (3)  is 
(4),  which  is  itself  the  obverse  of  (1). 

Similarly  the  converse  of  (2)  is  (4),  and  the  contraposite  of  (4)  is 
(3),  which  is  itself  the  obverse  of  (2). 

Of  the  four  theorems,  the  first  and  second  are  equivalent,  each  be- 
ing a  logical  consequence  of  the  other;  and  similarly  the  third  and 
fourth  are  equivalent. 

109.  Logical  independence.  Two  theorems  are  said  to  be  logically 
independent  when  neither  theorem  is  a  logical  consequence  of  the 
other ;  that  is,  when  neither  theorem  is  sufficient  to  prove  the  other, 
even  though  it  may  do  so  when  aided  by  some  fresh  geometric  fact. 

The  converse  theorems,  (1)  and  (3)  above,  are  logically  independent, 
neither  being  a  necessary  consequence  of  the  other  (103).  Similarly 
the  converse  theorems,  (2)  and  (4),  are  independent. 

Again,  the  obverse  theorems,  (1)  and  (4),  are  logically  independent, 
because  (1)  and  (3)  are  independent  and  (3)  is  equivalent  to  (4). 
Similarly  the  obverse  theorems,  (2)  and  (3),  are  independent. 

110.  The  three  transformations.  We  have  considered  three  pro- 
cesses of  transforming  a  known  theorem  ;  viz.  contraposition,  conver- 
sion, and  obversion.  Of  these,  only  the  first  is  a  process  of  logical 
inference.  In  the  other  two  cases  further  geometric  facts  must  be 
introduced  to  aid  the  inquiry  into  the  truth  or  falsity  of  the  two  new 
theorems.  It  is  only  necessary  to  examine  one  of  the  two,  however, 
for  they  are  logically  equivalent,  being  contraposites.  Hence  it  will 
never  be  necessary  to  demonstrate  more  than  two  of  the  four  related 
theorems,  care  being  taken  that  the  two  selected  are  not  contraposites  ; 
i.e.  the  two  selected  should  be  either  converse  or  obverse  to  each  other. 

Ex.  1.  In  which  of  the  theorems  1-20  is  the  converse  not  true  ? 
In  which  are  the  four  related  types  all  true  ?    Only  two  of  them  true  ? 

Ex.  2.  Show  that  91  has  three  converses.  [Here  the  hypothesis 
has  two  parts  ;  and  the  xjonclusion  may  be  interchanged  with  the 
whole  hypothesis,  or  with  either  of  the  parts.]  Which  one  of  these  is 
true  ?    State  the  three  converses  oi  '  when  .4  is  ^  and  C  is  D,  then  E 

is  f: 


PARALLEL  LINES  63 

PARALLEL  LINES 

The  previous  sections  have  treated  of  the  figures  formed 
by  two  or  three  intersecting  lines.  The  next  plane  figure 
to  be  considered  is  that  formed  by  two  indefinite  lines  that 
never  meet. 

111.  Definitions.  Two  straight  lines  lying  in  one  plane, 
which  will  never  meet  however  far  they  are  extended  both 
ways,  are  said  to  be  parallel. 

A  line  that  intersects  two  or  more  other  lines  (whether 
parallel  or  not)  is  said  to  be  a  transversal  to  those  lines. 

A  transversal  to  two  other  lines  forms  with  them  four 
interior  angles  and  four  exterior  angles. 

An  exterior  angle  and  the  non-adjacent  interior  angle  on 
the  same  side  of  the  transversal  are  called  corresponding 
angles. 

Two  non-adjacent  interior  angles  on  opposite  sides  of  the 
transversal  are  called  alternate  angles. 

In  the  figure  there  are  four  pairs  of  corresponding  angles 

(a,a'),  (6,6'),  (c,c'),  Kci'); 
cm 


and  there  are  two  pairs  of  alternate  angles 

(6,  c'),  (d,  a'). 

The  angles  made  with  two  lines  by  a  transversal  play  an 
important  part  in  the  theory  of  parallel  lines ;  this  will  be 
seen  in  the  following  theorem,  which  relates  to  alternate 
angles,  and  lays  a  foundation  for  the  succeeding  problem  on 
the  construction  of  parallel  lines. 


54  PLANE  GEOMETRY  —  BOOK  I 

Use  of  transversal  in  proving  parallelism, 

112.  Theorem  21.  If  two  lines  in  one  plane  are  such 
that  a  transversal  makes  a  pair  of  alternate  angles 
equal,  then  the  two  lines  are  parallel. 

Let  the  lines  iZ,'  and  MM'  be  cut  by  the  transversal  OP 
in  the  points  0,  P,  making  the  alternate  angles  LOP  and  M'PO 
equal  (fig.  1). 

L  o  L' 


MP  M' 

Fig.  1 


To  prove  that  the  lines  LL'  and  MM'  are  parallel. 

Suppose,  if  possible,  that  the  lines  are  not  parallel,  that 
is  to  say  that  they  meet  if  extended  far  enough  to  one  side 
or  the  other. 

Let  them  meet  in  the  point  Q  (fig.  2). 

Then  OPQ  is  a  triangle;  and  the  exterior  angle  LOP  is 
greater  than  the  interior  angle  OPQ  (79). 

This  is  contrary  to  the  hypothesis ;  hence  the  supposition 
made  is  false. 

Therefore  the  lines  LL'  and  MM*  do  not  meet,  however  far 
they  may  be  extended  both  ways ;  that  is  to  say  they  are 
parallel. 

State  the  contraposite  of  this  theorem,  and  compare  it  with  79. 

113.  Cor.  I.  If  two  lines  in  one  plane  are  such  that  a  trans- 
versal snakes  a  pair  of  corresponding  angles  equals  then  the  two 
lines  are  parallel. 

Ex.  1.  If  two  lines  are  cut  by  a  transversal,  and  if  two  interior 
angles  on  the  same  side  of  the  transversal  are  together  equal  to  a 
straight  angle,  then  the  two  lines  are  parallel. 

Ex.  2.  Two  lines  perpendicular  to  the  same  line  are  parallel. 


PARALLEL  LINES  66 

Drawing  a  parallel, 

114.   Pkoblem  8.     Through  a  given  point  to  draw  a 
line  parallel  to  a  given  line. 


y 

A. 


Outline.  Use  problem  7  (Art.  77),  and  prove  by  theorem  21 
(Art.  112). 

115.  Note.  This  solution  establishes  the  actual  existence  of 
parallel  straight  lines.  It  shows  that  there  is  at  least  one  line  pass- 
ing through  a  given  point  parallel  to  a  given  line.  The  possibility  of 
there  being  more  than  one  is  considered  in  the  next  article. 

Some  Properties  of  Parallels 

116.  The  parallel-postulate.  The  properties  of  parallel 
lines  rest  upon  a  fundamental  assumption,  usually  called 
the  postulate  of  parallels,  which  in  its  modern  form  may- 
be stated  as  follows : 

Let  it  he  granted  that  two  intersecting  straight  lines  cannot 
both  he  parallel  to  the  same  third  straight  line. 

This  is  equivalent  to  an  agreement  that  only  one  line  can 
be  drawn  through  a  given  point  parallel  to  a  given  line.  In 
other  words,  it  assumes  that  there  is  only  one  solution  to 
problem  8,  viz. : 

"  Through  a  given  point  to  draw  a  line  parallel  to  a  given 
line." 

The  truth  of  this  postulate  cannot  be  proved  from  any  of 
the  preceding  definitions  or  theorems.  It  may  be  regarded 
as  expressing  an  independent  property  of  Euclidean  space ; 
and  thus,  like  the  other  postulates,  it  has  the  nature  of  a 
definition. 

MoM.   ELEM.   GEOM.  6 


56  PLANE  GEOMETRY  —  BOOK  I 

From  the  parallel-postulate,  by  the  aid  of  theorem  21, 
the  following  theorem  is  derived,  which  is  the  converse  of 
theorem  21,  and  establishes  the  characteristic  property  of 
parallel  lines. 

Equality  of  alternate  angles, 

117.  Theorem  22.  If  two  parallels  are  cut  by  a 
transversals,  the  alternate  angles  are  equal. 

Let  the  parallels  LL\  MM*  be  met  by  the  transversal  OP. 


N' 


N 


M  P  M' 

To  prove  that  the  alternate  angles  LOP  and  M^PO  are  equal. 

Suppose,  if  possible,  that  one  of  them,  say  LOP,  is  the 
greater. 

Draw  the  line  NON^  cutting  off  from  the  greater  angle 
a  part  NOP  equal  to  the  less  angle  M^PO  (77). 

Then  NON'  is  parallel  to  MPM*  (112). 
Hence,  the  two  intersecting  lines  are  parallel  to  the  same 
line;  contrary  to  the  postulate  of  parallels. 

Therefore  the  supposition  fails;  that  is  to  say,  neither 
alternate  angle  is  greater  than  the  other ;  hence  the  alter- 
nate angles  are  equal. 

Why  could  117  not  be  derived  directly  from  112?  What  new 
geometric  fact  had  to  be  introduced  ?  Compare  the  statement  at  the 
end  of  103. 

Contraposite  of  theorem  22, 

118.  Cor.  I.  If  the  alternate  angles  are  not  equal,  the  lines 
are  not  parallel;  and  they  meet  at  that  side  of  the  transversal 
at  which  the  smaller  angle  lies. 


PARALLEL  LINES  57 

The  first  part  of  this  corollary  is  the  contraposite  of  theorem  22,  and 
is,  therefore,  true  by  106.  The  student  may,  as  an  exercise,  prove  it  by 
the  method  of  exclusion.  Compare  112,  118,  and  their  two  contra- 
posites.     How  is  the  latter  part  of  118  related  to  79  ? 

119.  Test  of  parallelism.  Arts.  112  and  118  together  show 
that  the  equality  of  the  alternate  angles  is  a  complete  test 
of  parallelism.  We  could  test  conclusively  whether  two 
lines  are  parallel  or  not  by  comparing  the  alternate  angles 
which  they  make  with  any  transversal.  If  the  angles  are 
equal,  the  lines  are  parallel ;  and  if  the  angles  are  not  equal, 
the  lines  are  not  parallel.  Such  a  complete  test  is  some- 
times called  "  a  necessary  and  sufficient  condition.^' 

120.  Sufficient  condition.  One  statement  is  said  to  be  a  sufficient 
condition  of  another  when  the  truth  of  the  first  is  sufficient  to  insure 
the  truth  of  the  second ;  or,  in  other  words,  when  the  truth  of  the 
second  is  a  necessary  consequence  of  the  truth  of  the  first. 

For  instance,  when  we  say  that  the  statement  "  ^  is  5  "  is  a  suf- 
ficient condition  of  the  statement  "  O  is  D"  we  mean  only  that 

When  A  is  B,  0  is  D: 

Thus  Art.  112  asserts  that  the  equality  of  the  alternate  angles  is  a 
sufficient  condition  of  parallelism. 

121.  Necessary  condition.  One^statement  is  said  to  be  a  neces- 
sary condition  of  another  when  the  second  cannot  be  true  unless  the 
first  is  true  ;  or,  in  other  words,  when  the  opposite  of  the  second  is  a 
necessary  consequence  of  the  opposite  of  the  first. 

For  instance,  when  we  say  that  the  statement  "^  is  ^"  is  a  neces- 
sary condition  of  the  statement  "  O  is  D"  we  mean  only  that 

When  A  is  not  J5,  C  is  not  D. 

Thus  Art.  118  asserts  that  the  equality  of  the  alternate  angles  is  a 
necessary  condition  of  parallelism. 

Arts.  112  and  118  together  show  that  the  equality  of  the  alternate 
angles  is  both  a  necessary  and  a  sufficient  condition  of  parallelism. 

Ex.  1.  If  one  statement  is  a  sufficient  condition  of  another,  show 
that  the  latter  is  a  necessary  condition  of  the  former. 

Ex.  2.  If  one  statement  is  a  necessary  condition  of  another,  show 
that  the  latter  is  a  sufficient  condition  of  the  former. 


58  PLANE  GEOMETRY  —  BOOK  I 

Ex.  3.  If  one  statement  is  a  necessary  and  suflBcient  condition  of 
another,  then  the  latter  is  a  necessary  and  sufficient  condition  of  the 
former. 

Ex.  4.  Show  that  the  statement  "  X  equals  F"  is  a  sufficient  (but 
not  a  necessary)  condition  of  the  statement  "  X  is  greater  than  half 
F"  ;  and  that  the  latter  is  a  necessary  (but  not  a  sufficient)  condition 
of  the  former. 

Equality  of  corresponding  angles, 

122.  Cor.  2.  If  two  parallels  are  cut  by  a  transversal,  the 
corresponding  angles  are  equal  ;  and  conversely. 


Interior  angles  supplemental, 

123.  Cor.  3.  If  tico  parallels  are  cut  by  a  transversal j 
the  two  interior  angles  on  the  same  side  are  together  equal  to  a 
straight  angle  ;  and  conversely. 

Interior  angles  not  supplemental, 

124.  Cor.  4.  If  two  lines  are  cut  by  a  transversal^  and  if 
the  interior  angles  on  the  same  side  of  the  transversal  are 
together  less  than  a  straigfU  angle,  then  the  lines  will  meet  at 
that  side  of  the  transversal  at  which  these  angles  are.* 

What  relation  does  124  bear  to  80  ?  What  relation  does  124  bear 
to  each  of  the  parts  of  123  ?  Which  of  these  four  theorems  are  logical 
equivalents  ?    Which  of  them  are  not  logical  equivalents  ? 

*  The  statement  in  corollary  4  was  adopted  as  a  postulate  by  Euclid, 
and  was  placed  at  the  foundation  of  his  theory  of  parallels.  The 
form  stated  in  116  was  first  given  in  Playfair's  edition  of  Euclid's 
'Elements  of  Geometry'  (1813),  and  has  been  generally  adopted  by 
modern  writers.  An  earlier  suggestion  of  this  form  of  the  postulate  is 
found  in  '  Rudiments  of  Mathematics,'  by  W.  Ludlam,  St.  John's 
College,  Cambridge  (1794). 


PARALLEL  LINES  59 

Another  test  of  parallelism.  Arts.  123  and  124  furnish  a 
useful  test  as  to  whether  two  lines  are  parallel  or  not  (com- 
pare 119).  Art.  124  is  often  used  to  prove  that  certain  lines 
will  meet  if  prolonged  in  a  certain  way. 

125.  Cor.  5.  Lines  that  are  parallel  to  the  same  line  are 
parallel  to  each  other.     (Use  122.) 

Ex.  A  line  perpendicular  to  one  of  two  parallels  is  perpendicular 
to  the  other.  ' 

Angles  having  parallel  sides, 

126.  Theorem  23.  If  two  angles  have  the  two  sides  of 
one  respectively  parallel  to  the  two  sides  of  the  other 
( parallel  lines  being  at  the  same  side  of  the  line  joining 
the  vertices  of  the  angles),  then  the  angles  are  equal. 

Let  AOB,  a'o'b'  be  the  angles  having  OA  and  0'^'  parallel 
and  on  the  same  side  of  00' ;  and  similarly  for  OB  and  O'B'. 


To  prove  that  the  angles  AOB  and  A'O'b'  are  equal. 
Let  O'A'  meet  OB  (extended  if  necessary)  in  C. 
The  angles  AOB,  A'O'b'  are  each  equal  to  A'CB  (122). 
Hence,  they  are  equal  to  each  other. 

127(a).  Cor.  i.  If  tivo  angles  have  the  two  sides  of  one 
respectively  parallel  to  the  two  sides  of  the  other  (parallel  lines 
being  at  opposite  sides  of  the  lirie  joining  the  vertices  of  the 
angles),  then  the  angles  are  equal. 


60  PLANE  GEOMETRY  —  BOOK  I 

127  (b).  Cor.  2.  If  two  angles  have  the  two  sides  of  one  re- 
spectively parallel  to  the  tivo  sides  of  the  other  {two  of  the 
parallels  being  at  the  same  side  of  the  line  joining  the  vertices 
and  the  other  two  being  on  opposite  sides  of  that  line),  then  the 
angles  are  supplemental. 

Theory  op  Parallels  applied  to  Angle-sums 

The  following  twO  theorems,  with  their  inferences,  illus- 
trate how  the  theory  of  parallels  may  be  used  in  the  addi- 
tion and  subtraction  of  angles. 

Sum  of  two  angles  of  a  triangle, 

128.  Theorem  24.  When  any  side  of  a  triangle  is 
extended,  the  exterior  angle  is  equal  to  the  sum  of  the 
two  interior  opposite  angles. 

Let  the  side  AC  of  the  triangle  ABChQ  extended  to  E, 


To  prove  that  the  exterior  angle  BCE  is  equal  to  the  sum 
of  the  opposite  interior  angles  CAB  and  CBA. 

Draw  the  line  CD  parallel  to  AB  (114). 

The  angles  BCD  and  CAB  are  equal  (122). 
Also  the  angles  DCB  and  CBA  are  equal  (117). 
Therefore,  by  addition  of  equals,  the  angle  ECB  is  equal 
to  the  sum  of  the  angles  CAB  and  CBA. 

Ex.  1.  When  two  lines  are  met  by  a  transversal,  the  difference  of 
two  corresponding  angles  is  equal  to  the  angle  between  the  two  lines. 

Ex.  2.  The  difference  between  two  alternate  angles  is  equal  to 
what  angle  ? 


PARALLEL  LINES  61 

By  how  much  does  the  sum  of  the  two  interior  angles  at  one  side  of 
a  transversal  exceed  the  sum  of  the  interior  angles  at  the  other  side  ? 

Ex.  3.  In  the  figure  of  Art.  87,  prove  that  the  angle  ADB  equals 
half  ACB. 

Ex.  4.  In  the  figure  of  Art.  81,  prove  that  the  difference  of  the 
angles  ABG  and  ACB  equals  double  DBG. 

[Angle  ABC  equals  the  sum  of  ABD  and  DBC,  which  equals  the 
sum  of  ADB  and  DBC,  etc.] 

Angle-sum  in  a  triangle, 

129.  Theorem  25.  The  sum  of  the  three  interior 
angles  of  a  triangle  is  equal  to  a  straight  angle. 

[Use  the  equality  proved  in  128  ;  and  add  to  both  members  the 
third  interior  angle.] 

Ex.  1.  In  a  right  triangle  the  acute  angles  are  complemental ;  and 
in  an  isosceles  right  triangle  the  acute  angles  are  each  equal  to  half  a 
right  angle. 

Ex.  2.  In  any  isosceles  triangle  each  of  the  equal  angles  is  equal  to 
the  complement  of  half  the  third  angle. 

Ex.  3.  In  an  equilateral  triangle  each  angle  is  equal  to  two  thirds 
of  a  right  angle. 

Ex.  4.  Show  how  to  construct  an  angle  equal  to  one  third  of  a 
right  angle.     Hence  show  how  to  trisect  a  given  right  angle. 

Ex.  5.     Trisect  a  given  straight  angle. 

130.  Cor.  If  tivo  triangles  have  two  angles  of  one  equal  to 
two  angles  of  the  other,  then  the  third  angles  are  equal. 

EXERCISES 

1.  Through  a  given  point  draw  a  line  making  with  a  given  line  an 
angle  equal  to  a  given  angle. 

2.  If  two  lines  are  respectively  perpendicular  to  two  other  lines,  the 
angles  formed  by  the  first  pair  are  respectively  equal  to  the  angles 
formed  by  the  second  pair. 

3.  Two  lines  perpendicular  to  two  parallel  lines,  respectively,  are 
parallel. 


62  PLANE  GEOMETRY  —  BOOK  I 


CONSTRUCTION  OF  TRIANGLES 

131.  There  is  a  large  class  of  problems  involving  the 
construction  of  triangles  to  satisfy  certain  prescribed  con- 
ditions involving  the  sides  and  angles. 

In  a  triangle  the  three  sides  and  the  three  angles  are 
called  its  six  parts.  For  convenience  the  sides  will  be 
denoted  by  a,  6,  c  and  the  opposite  angles  respectively  by 
A,  Bj  a  In  a  right  triangle  the  side  opposite  the  right 
angle  is  called  the  hypotenuse. 

The  simplest  condition  that  can  be  imposed  on  the  con- 
struction of  a  triangle  is  the  assignment  of  certain  line 
segments  or  angles  to  which  some  of  the  six  parts  are  to 
be  made  equal.  Those  parts  which  are  to  be  made  equal 
to  prescribed  segments  or  angles  are  said  to  be  given  or 
known,  and  the  remaining  parts,  about  which  nothing  is 
prescribed,  are  said  to  be  unknown. 

In  each  of  the  five  following  problems  three  of  the  six 
parts  are  given,  and  it  is  required  to  find  by  a  geometric 
construction  a  triangle  answering  to  the  prescribed  con- 
ditions, and  incidentally  to  determine  the  three  unknown 
parts  of  the  triangle. 

The  solution  of  such  a  problem  has  three  divisions : 

(1)  To  make  the  actual  construction  by  means  of  processes 
that  ultimately  involve  only  the  drawing  of  straight  lines 
and  of  circular  arcs. 

(2)  To  prove  by  the  use  of  previous  propositions  that  the 
figure  so  constructed  satisfies  the  prescribed  conditions. 

(3)  To  discuss  the  solution ;  i.e.  to  examine  what  limita- 
tions there  are  on  the  data  so  that  it  may  be  possible  to 
satisfy  the  demands  ;  and  under  what  circumstances  it  will 
be  possible  to  satisfy  them  in  only  one  way,  or  in  more 
than  one  way ;  and  also  to  examine  certain  special  and 
limiting  cases. 

When  the  data  are  such  that  the   demands  cannot  be 


CONSTRUCTION  OF  TRIANGLES  63 

satisfied  by  any  triangle,  the  problem  is  said  to  have  no 
solution. 

If  they  can  be  satisfied  by  one,  and  only  one,  type  of 
triangle  (i.e.  if  all  the  triangles  fulfilling  the  stated  con- 
ditions can  be  superposed),  the  problem  is  said  to  have 
a  unique  solution. 

If  the  demands  can  be  satisfied  by  two  or  some  greater 
definite  number  of  distinct  types  of  triangles  not  capable  of 
superposition,  the  problem  is  said  to  have  a  determinate 
but  ambiguous  solution. 

If  the  demands  can  be  satisfied  by  an  indefinite  number 
of  triangles,  the  problem  is  said  to  have  an  indeterminate 
solution. 

To  prepare  the  way  for  the  solution  it  is  usually  best  to 
make  a  preliminary  analysis  of  each  problem.  This  may 
be  described  in  a  general  way  as  follows : 

Suppose  the  problem  solved,  and  the  required  figure 
drawn.  Mark  the  parts  that  are  supposed  to  be  equal  to 
"given"  lines  or  angles.  Analyze  the  figure  to  discover 
the  relations  of  the  known  and  unknown  parts.  Draw  any 
lines  that  may  help  to  bring  them  into  closer  relations  to 
each  other.  Observe  which  of  the  problems  already  solved 
could  be  used  to  construct  the  various  parts  subject  to  the 
given  conditions.  This  is  called  "  reducing  the  problem  to 
previous  ones." 

After  this  reduction  (or  analysis)  has  been  made,  perform 
these  simpler  constructions  in  their  proper  order,  thus  build- 
ing up  the  figure  step  by  step.  Such  building-up  process  is 
called  a  synthesis.  Then  will  follow  the  proof,  and  the 
discussion  as  already  stated. 

In  many  cases  the  preliminary  analysis  is  so  simple  that 
it  will  not  be  given,  but  the  student  should  always  make 
such  analysis  before  consulting  the  synthetic  solution. 

It  is  advisable  to  make  some  of  the  actual  constructions 
with  ruler  and  compasses;  and  it  affords  better  geometrical 
training  to  dispense  with  all  other  mechanical  aids. 


64  PLANE  GEOMETRY  —  BOOK  I 

Three  sides  given, 

132.   Problem  9.     To  construct  a  triangle  having  its 
sides  equal  to  three  given  lines. 

Let  a,  h,  c,  be  the  three  given  lines. 

c  V 


Draw  any  line  equal  to  a.  With  its  ends  as  centers,  and 
radii  equal  to  6,  c  respectively,  describe  circles.  Join  their 
point  of  intersection  to  the  ends  of  a. 

The  triangle  so  formed  fulfills  the  given  conditions,  since 
it  has  its  sides  respectively  equal  to  the  given  lines. 

Discussion.  The  limitations  on  the  data,  in  order  that  a 
triangle  satisfying  the  demands  may  exist,  are  that  the  sum 
of  any  two  of  the  given  segments  must  be  greater  than  the 
third  (87). 


/'c 


\ 


Show  that  this  appears  in  the  actual  process  of  construction.  (In 
the  figure  the  sum  of  a  and  c  is  less  than  6,  and  the  circles  do  not  in- 
tersect.) Examine  the  limiting  cases  in  which  the  sum  of  any  two 
of  the  given  lines  is  equal  to  the  third. 

If  the  other  intersection  of  the  two  circles  had  been  selected  ;  or  if 
the  side  h  had  been  taken  first,  and  with  its  ends  as  centers,  circles 
had  been  described  with  radii  a  and  c ;  the  different  triangles  so  obtained 
could  all  be  superposed  upon  the  first  by  suitable  movement  (66). 

Thus  there  is  a  unique  solution  when  each  of  the  given  segments  is 
less  than  the  sum  of  the  other  two  ;  and  if  any  of  these  conditions  be 
violated  there  is  no  solution. 


CONSTRUCTION  OF  TRIANGLES  '        66 

Given  two  sides  and  the  included  angle, 
133.   Problem  10.     To  construct   a  triangle  having 
two  of  its  sides  equal  to  two  given  lines,  and  the  in- 
cluded angle  equal  to  a  given  angle. 

Let  a,  6  be  the  given  lines  and  C  the  given  angle.  - 


B 

On  any  line  lay  off  a  segment  CB  equal  to  a.  Through 
one  extremity  C  of  this  segment  draw  a  line  making  an 
angle  equal  to  the  given  angle  C.  On  this  line  lay  off  from 
C  a  segment  CA  equal  to  b,  and  join  AB. 

The  triangle  so  formed  has  the  prescribed  parts. 

Discussion.  Show  that  the  only  limitation  on  the  data  is 
that  the  given  angle  must  be  less  than  a  straight  angle  (129). 

Show,  as  in  132,  that  the  solution  is  unique  (64). 

Criven  two  angles  and  the  included  side, 

134.  Problem  11.  To  construct  a  triangle  having 
two  of  its  angles  equal  to  two  given  angles,  and  the 
included  side  equal  to  a  given  line. 


Make  two  applications  of  77. 

What  limitation  is  there  on  the  sum  of  the  two  given 
angles  (129)  ? 

Show  that  the  solution  is  unique. 

Ex.  Construct  a  right  triangle  being  given  one  of  the  sides  form- 
ing the  right  angle  and  either  of  the  acute  angles. 


66         '  PLANE  GEOMETRY  —  BOOK  I 

Given  two  angles  and  the  side  opposite  one, 

135.  Problem  12.  To  construct  a  triangle  having 
two  of  its  angles  equal  to  two  given  angles,  and  the 
side  opposite  to  a  specified  one  of  these  angles  equal 
to  a  given  line. 

Let  Ay  B  be  the  given  angles  and  a  the  given  line. 


First  add  the  angles  A  and  B  together,  and  subtract  the 
sum  from  a  straight  angle.  (This  can  be  conveniently 
done  by  taking  any  line,  then  at  any  point  O  constructing 
an  angle  equal  to  A,  and  an  adjacent  angle  equal  to  B.) 

The  remaining  part  of  the  straight  angle  is  equal  to  the 
third  angle,  C,  of  the  required  triangle  (129). 

Then  in  the  triangle  ABC  the  angles  B,  C  are  known,  and 
the  included  side  a.  Hence  the  triangle  can  be  constructed 
as  in  134. 

Discussion.  When  is  there  no  solution  (129)  ?  Is  there 
any  ambiguity  (65,  95)  ? 

Ex.  1.  Construct  an  isosceles  triangle,  being  given  its  base  and 
opposite  angle  (129). 

Ex.  2.  Construct  a  right  triangle,  being  given  the  hypotenuse  and 
one  acute  angle. 

Ex.  3.  Draw  a  line  parallel  to  the  base  of  a  triangle  so  that  the 
portion  intercepted  between  the  sides  may  be  equal  to  a  given  line. 

Given  two  sides  and  the  angle  opposite  one, 
136.   Problem  13.     To  construct  a  triangle  having 
two  of  its  sides  equal  to  two  given  lines,  and  the 


CONSTRUCTION  OF  TRIANGLES  67 

angle  opposite  a  specified  one  of  these  sides  equal  to 

a  given  angle. 

Let  a,  h  be  the  given  lines  and  A  the  given  angle. 


In  any  line  AB  take  any  point  A ;  and  draw  AC  making 
the  angle  BAC  equal  to  the  given  angle  (77).  Lay  off  AC 
equal  to  that  one  of  the  given  segments  which  is  to  be  equal 
to  the  side  adjacent  to  the  given  angle.  With  C  as  center 
and  radius  equal  to  the  other  given  segment  draw  a  circle 
cutting  the  line  AB  in  the  point  B ;  and  join  CB. 

The  triangle  so  formed  evidently  fulfills  the  stated  con- 
ditions. 

Discussion.  The  discussion  of  this  problem  falls  into 
three  divisions  according  to  the  species  of  the  given  angle, 
and  each  division  has  a  number  of  cases  according  to  the 
greater  or  less  magnitude  of  the  two  given  segments.  An 
examination  of  the  different  cases  will  be  found  very  in- 
structive. 

I.   Let  the  given  angle  A  be  acute. 

(1)  In  the  process  of   construction  let  the 
segment  a  be  less  than  the  perpendicular  p 
drawn  from  C  to  AB. 

No  solution  is  possible  (85). 

(2)  Let  a  equal  p. 

There  is  one  and  only  one  triangle  answer- 
ing the  conditions  ;  because  any  triangle  having 
the  specified  parts  equal  to  a,  b,  A,  is  right 
angled,  and  is  therefore  superposable  on  the 
triangle  ABC  (98). 


68 


PLANE  GEOMETRY  —  BOOK  I 


(3)  Let  a  be  greater  than  p  and  less  than  b. 
The  arc  meets  AB  in  two  points  B,  B\  both 

situated  on  that  portion  of  the  line  AB  which 
forms  one  side  of  the  given  angle  BAC.  Hence 
there  are  two  triangles  ABC,  AB'C,  each  having 
the  specified  parts  equal  to  a,  b,  A. 

The  other  parts  c,  B,  C  and  c',  B\  C'  are  respectively 
unequal  in  the  two  triangles.  In  particular  the  angle  B'  in 
the  triangle  AB'c  is  the  supplement  of  the  angle  B  in  the 
triangle  ABC. 

The  stated  conditions  are  not  satisfied  by  any  other  type 
of  triangle ;  because  any  other  triangle  having  the  specified 
parts  equal  to  a,  6,  A,  would  have  the  angle  opposite  the 
side  b  equal  either  to  the  angle  B  or  to  its  supplement  (97), 
and  is  therefore  superposable  on  one  or  the  other  of  the 
triangles  ABC,  AB'c  (65). 

Thus  there  are  two  and  only  two  solutions. 

(4)  Let  a  equal  b. 
The  point  B'  coincides  with  A.     Show  that 

the  solution  is  unique  (64,  65). 

(5)  Let  a  exceed  b. 
Show  that  B,  B'  fall  at  opposite 

sides  of  A,  and  that  the  solution 


IS  unique. 

II.   Let  the  given  angle  A 


B'    A 
be  a  right  angle. 

(1)  Let  a  be  less  than  6,  or  a  equal  6.     No  triangle. 

(2)  Let  a  exceed  b. 

The  points  5,  B'  are  at  opposite  sides  of  A;   and  the 
triangles  ABC,  AB'c  each  answer  the  conditions. 


CONSTBUCTION  OF  TRIANGLES  69 

Show  that  these  triangles  are  superposable ;  that  so  are 
all  other  triangles  fulfilling  the  stated  conditions  j  and  that 
the  sohition  is  then  unique. 

III.   Let  the  given  angle  A  be  obtuse. 

(1)  Let  a  be  less  than  b,  or  a  equal  b.     No  solution. 
Show  that  this  conclusion  could  also  be  drawn  from  the 

fact  that  there  cannot  be  two  obtuse  angles  in  a  triangle  (80). 

(2)  Let  a  exceed  b. 

The  points  B,  b'  are  on  opposite  sides  of  A,  and  the  triangle 
AB'c  is  not  a  solution. 


Any  other  triangle  having  the  specified  parts  equal  to  a, 
6,  A,  would  have  the  angle  opposite  b  equal  either  to  B  or  to 
its  supplement  (97).  In  the  former  case  the  triangle  would 
be  superposable  on  ABC.  In  the  latter  case  the  triangle 
could  not  exist  since  the  supplement  of  B  is  obtuse,  and 
there  cannot  be  two  obtuse  angles  in  a  triangle. 

Hence  there  is  but  one  type  of  triangle  answering  the 
requirements,  and  the  solution  is  unique. 

Ex.  1.  Give  a  summary  of  the  limitations  there  are  on  the  data  in 
order  that  any  solution  may  be  possible. 

Ex.  2.  Summarize  the  circumstances  under  which  there  is  a  unique 
solution. 

Ex.  3.  When  is  the  solution  ambiguous  ?  Quote  the  previous 
theorem  from  which  it  is  inferred  that  there  can  be  no  third  type  of 
triangle  fulfilling  the  conditions. 

137.  Determinate  and  indeterminate  solutions.  It  has  per- 
haps been  noticed  that  the  assigned  conditions  in  these  five 
problems  of   construction  of  a  triangle  correspond  respec- 


70  PLANE  GEOMETRY  —  BOOK  I 

tively  to  the  five  conditions  of  equality  of  two  triangles  (64- 
66,  95,  97) ;  and  that  the  determinateness  of  the  solution 
is  in  each  case  tested  by  applying  the  corresponding  condition 
of  equality. 

For  instance,  the  solution  of  134  is  uniquely  determinate, 
because  any  two  triangles  that  have  two  sides  and  the  in- 
cluded angle  in  each  respectively  equal,  are  superposable  by 
the  first  condition  of  equality. 

Again,  the  solution  of  136  is  in  one  case  ambiguously 
determinate,  because  any  two  triangles  that  have  two  sides 
and  the  angle  opposite  one  of  them  in  each  equal  have  the 
angles  opposite  the  other  equal  sides  either  equal  or  supple- 
mental, and  because  there  is  (in  the  case  referred  to)  no  way 
of  deciding  the  species  of  the  angle  in  question  from  previous 
principles. 

In  each  of  these  problems  three  parts  were  given.  If 
only  two  parts  are  assigned,  an  indefinite  number  of  distinct 
types  of  triangles  answering  the  requirements  can  be  con- 
structed. 

For  instance,  if  only  two  sides  are  given,  the  included 
angle  can  be  assumed  at  pleasure,  and  an  indefinite  number 
of  distinct  triangles  can  be  formed  so  as  to  have  the  given 
sides. 

Similarly  if  only  two  angles  are  given,  the  adjacent  side 
can  be  assumed  arbitrarily,  and  the  solution  is  indeter- 
minate. 

The  three  angles,  however,  do  not  constitute  three  inde- 
pendent data;  for  then  nothing  more  is  given  than  when 
only  two  angles  are  assigned  (as  the  third  angle  could  be 
found  by  subtracting  the  sum  of  the  first  two  from  a  straight 
angle). 

Thus  the  problem  to  construct  a  triangle  having  its  three 
angles  equal  to  three  assigned  angles,  is  either  impossible  or 
indeterminate;  impossible  if  the  sum  of  the  three  given 
angles  is  not  equal  to  a  straight  angle;  indeterminate  if 
it  is. 


CONSTRUCTION  OF  TRIANGLES  71 


EXERCISES 

1.  Given  base,  vertical  angle,  and  sum  of  sides,  construct  the 
triangle. 

Outline,  In  the  figure  of  Art.  87,  in  which  AD  is  the  sum  of  the 
sides  AC  and  CB,  show  that  the  angle  ADB  is  half  the  vertical  angle 
ACB  ;  and  hence  that  the  triangle  ABB  can  be  constmcted  from  the 
data  (136).     Then  show  how  to  construct  the  triangle  ABC. 

2.  Given  base,  vertical  angle,  and  difference  of  sides,  construct  the 
triangle. 

Outline.  In  the  figure  of  Art.  81,  in  which  CD  is  the  difference  of 
sides,  show  that  the  angle  CDB  equals  the  sum  of  the  vertical  angle 
and  half  its  supplement.  Construct  the  triangle  CDB  by  136,  and 
then  the  triangle  ABC. 

3.  Given  base,  difference  of  sides,  and  difference  of  base  angles, 
construct  the  triangle. 

[Use  128,  ex.  4.] 

4.  Given  base,  an  adjacent  angle,  and  the  sum  (or  difference)  of 
the  other  two  sides,  construct  the  triangle. 

5.  Given  the  angles  and  the  perimeter,  construct  the  triangle. 
Analysis.     Prolong  base  BC  both  ways,  making  BB'  equal  BA,  and 

CC  equal  CA.    Prove  B'  C  equal  to  perimeter ;  and  angle  B'  equal 
half  B,  etc.     Give  synthesis. 

6.  Given  the  angles  and  the  sum  (or  difference)  of  two  sides,  con- 
struct the  triangle. 

7.  In  the  figure  of  Art.  81,  prolong  CA  until  AE  equals  AB,  and 
draw  EB;  prove  that  angle  DBE  equals  the  sum  of  DEB  and  BDE, 
and  is  a  right  angle. 

8.  Construct  a  triangle,  being  given  the  base,  the  sum  of  sides, 
and  the  difference  of  the  base  angles.    [Use  ex.  7.] 

9.  Construct  an  equilateral  triangle  such  that  the  perpendicular 
from  the  vertex  to  the  base  may  be  equal  to  a  given  line. 


McM.  ET.EM.  GEOM. 


72  PLANE  GEOMETRY  —  BOOK  I 


QUADRANGLES 

Attention  has  hitherto  been  given  to  various  properties 
of  the  plane  figures  formed  by  two  or  three  straight  lines. 
The  figure  that  next  presents  itself  is  that  formed  by  four 
lines  each  of  which  meets  the  next  one  in  order. 

138.  Definitions.  A  plane  figure  formed  by  four  line-seg- 
ments that  inclose  a  portion  of  the  plane  surface  is  called 
a  quddrilatercuL  figure,  or  a  quadrangle. 

These  line-segments  are  called  the  sides,  and  their 
extremities  the  vertices  of  the  quadrangle. 

The  angles  formed  by  adjacent  sides,  and  situated  toward 
the  interior  of  the  boundary,  are  called  the  interior  angles 
of  the  quadrangle,  or  briefly  the  angles. 


The  exterior  angles  conjunct  to  these  will  be  called  for 
brevity  the  conjunct  angles. 

A  concave  angle  formed  by  one  side  and  the  prolongation 
of  an  adjacent  side  is  called  an  exterior  angle. 

If  all  of  the  conjunct  angles  are  convex  (21),  the  quadrangle 
is  called  convex. 

If  one  of  the  conjunct  angles  is  concave,  the  figure  is  said 
to  be  concave  at  that  angle. 

In  a  convex  quadrangle  all  the  interior  angles  are  concave, 
and  no  side  when  prolonged  traverses  the  figure ;  but  a  con- 
cave quadrangle  has  one  of  the  interior  angles  convex,  and 
the  sides  of  this  angle  traverse  the  figure  when  prolonged. 


QUADRANGLES  73 

A  line  connecting  two  non-adjacent  vertices  is  called  a 
diagonal. 

The  sum  of  the  sides  is  called  the  perimetery  and  the 
sum  of  the  angles  the  angle-sum. 

In  a  convex  quadrangle  the  sum  of  the  exterior  angles 
formed  by  prolonging  each  side  one  way,  no  two  adjacent 
sides  being  prolonged  through  the  same  vertex,  is  called  the 
exterior  angle-sum. 

The  four  sides  and  four  angles  are  called  the  eight  parts 
of  the  quadrangle. 

JPrimary  relations  of  parts, 

139.  Eelation  1.  The  suin  of  any  three  sides  is 
greater  than  tlw  fourth  (89). 

140.  Cor.  The  sum  of  any  two  sides  is  greater  tJian  the 
difference  of  the  other  two. 

141.  Eelation  2.  The  angle-sum  is  equal  to  a 
perigon. 

[Divide  the  quadrangle  into  two  triangles  by  a  diagonal  and  apply 
129.] 

142.  Cor.  I.  A  conjunct  angle  is  equal  to  the  sum  of  the 
three  non-adjacent  interior  angles. 

143.  Cor.  2.  Only  one  of  the  interior  angles  in  a  quad- 
rangle can  he  convex. 

144.  Cor.  3.  If  two  quadrangles  have  three  angles  of  one 
equal  respectively  to  three  angles  of  the  other,  the  remaining 
angles  are  equal. 

Ex.  1.  The  sum  of  the  four  sides  of  a  quadrangle  is  greater  than 
the  double  of  either  diagonal,  and  greater  than  the  sum  of  the  diago- 
nals. 

Ex.  2.  The  sum  of  the  four  interior  angles  is  equal  to  one  third  of 
the  sum  of  the  four  conjunct  angles. 


74 


PLANE  GEOMETR  Y  —  BOOK  I 


Some  Conditions  of  Equality 

The  following  three  theorems  relate  to  the  equality  of  two 
quadrangles  under  certain  conditions. 

Three  sides  and  two  included  angles, 

145.  Theorem  26.  If  two  quadrangles  have  three 
sides  and  the  two  included  angles  of  one  equal  to  the 
corresponding  parts  in  the  other,  the  figures  are  equal. 


Outline.  Superpose  the  equal  parts,  and  show  that  the  coincidence 
of  the  remaining  parts  will  follow  as  in  (>4. 

Tivo  adjacent  sides  and  three  angles. 

146.  Theorem  27.  If  two  quadrangles  have  two  ad- 
jacent sides  and  any  three  angles  of  one-  equal  to  the 
corresponding  parts  in  the  other,  the  figures  are  equal. 


Outline.  The  remaining  angles  are  equal  (144).  Superpose  the 
equal  parts,  and  show  that  the  remaining  parts  will  coincide. 

Tivo  opposite  sides  and  three  angles, 

147.  Theorem  28.  If  two  quadrangles  have  two  op- 
posite sides  and  any  three  angles  of  one  equal  to  the 
corresponding  parts  in  the  other,  the  figures  are 
equal. 


QUADRANGLES 


75 


In  the  quadrangles  ABCD  and  j'b'c'd',  let  AB  equal  A'B', 
CD  equal  C'D'.  Also  let  angle  A  equal  A',  B  equal  ^',  c  equal 
C',  and  consequently  jD  equal  D'  (144). 


To  prove  that  the  quadrangles  are  equal. 

Prolong  BC  and  AD  to  meet  in  P;  also  5'c'  and  A'd'  to 
meet  in  P'. 

The  angle  PCZ)  equals  p'c'd',  and  Ci)P  equals  (7'i>'P'. 

Therefore  the  triangles  PCD  and  p'c'd',  having  a  side  and 
the  two  adjacent  angles  in  each  equal,  are  themselves  equal. 

For  a  similar  reason  the  triangle  PDA  equals  p'b'a'. 

By  subtraction  of  equals  from  equals  the  line  BC  equals 
B'C',  and  AD  equals  A'd'. 

Hence  the  quadrangles  are  equal  by  the  preceding  theorem. 


Special  Kinds  of  Quadrangles 

148.  Definitions.  A  quadrangle  that  has  a  pair  of  its  op- 
posite sides  parallel,  and  the  other  pair  not  parallel,  is 
called  a  trapezoid.  One  that  has  both  pairs  of  opposite 
sides  parallel  is  a  parallelograwb. 


In  contradistinction  a  quadrangle  that  has  neither  pair  of 
sides  parallel  is  called  a  trapezium. 


76  PLANE  GEOMETRY  —  BOOK  I 

A  trapezoid  whose  opposite  non-parallel  sides  are  equal  is 
said  to  be  isosceles. 


A  parallelogram  that  has  two  adjacent  sides  equal  is  called 
a  rhonibus. 

[It  is  shown  later  (154)  that  the  four  sides  of  a  rhombus  are  equal.] 

A  parallelogram  that  has  one  of  its  angles  right  is  called 
a  rectangle. 

[It  will  appear  (150)  that  all  the  angles  of  a  rectangle  are  right 
angles.] 


A  rectangle  that  has  two  adjacent  sides  equal  is  a  square. 

[It  is  shown  later  (154)  that  all  the  sides  of  a  square  are  equal.  A 
square  is  at  once  a  rhombus  and  a  rectangle.] 

PARALLELOGRAMS   AND   TRAPEZOIDS 

Theorems  29-36  with  their  corollaries  establish  the  prin- 
cipal properties  of  parallelograms  and  trapezoids. 

Angles  of  a  parallelogram. 

149.  Theorem  29.  Any  two  consecutive  angles  of  a 
parallelogram  are  supplemental;  and  any  two  oppo- 
site angles  are  equal. 

[Apply  126  and  corollaries,  or  prove  independently  by  the  method 
of  that  article.] 


QUADRANGLES  77 

150.  Cor.     All  the  angles  of  a  rectangle  are  right  angles. 

Ex.  Show  that  in  a  trapezoid  the  sum  of  a  certain  pair  of  angles  is 
equal  to  the  sum  of  the  other  pair. 

Converse  of  last  theorem, 

151.  Theorem  30.  A  quadrangle  that  has  both  pairs 
of  opposite  angles  equal  is  a  parallelogram. 

Outline.  The  sum  of  two  consecutive  angles  is  equal  to  the  sum  of 
the  other  two,  and  is  therefore  equal  to  a  straight  angle  (141)  ;  hence 
the  opposite  sides  are  parallel  (113). 

Ex.     State  and  prove  the  converse  of  the  first  part  of  theorem  29. 

152.  Cor.  If  the  sum  of  one  pair  of  angles  of  a  quadrangle 
is  equal  to  the  sum  of  the  other  pair,  the  figure  is  a  trapezoid. 

Sides  of  a  parallelogram, 

153.  Theorem  31.  In  a  parallelogram  the  opposite 
sides  are  equal,  and  a  diagonal  divides  the  figure 
into  two  equal  triangles. 

Let  ABCD  be  a  parallelogram,  and  AC  its  diagonal.  • 


To  prove  that  AB  equals  CD,  EC  equals  AD,  and  that  the 
triangle  ABC  equals  ACD. 

Comparing  the  triangles  ABC  and  CDA,  the  angles  BAC 
and  ACD  are  equal  (117);  the  angle  BCA  equals  DAC;  and 
the  side  ^C  is  common ;  therefore  the  triangles  are  equal 
(65),  and  hence  AB  equals  CD,  and  BC  equals  AD. 

154.  Cor.  I.  All  the  sides  of  a  rhombus  are  equal  ;  and  so 
are  all  the  sides  of  a  square. 


78  PLANE  GEOMETRY  —  BOOK  I 

155.  Cor.  2.     Parallel    lines    intercept  equal   segments  on 
parallel  lines. 

Ex.  1.     Show  how  to  construct  a  square  on  a  given  line  as  side. 

Ex.  2.     Show  how  to  construct  a  square  on  a  given  line  as  diagonal. 

Ex.  3.     To  construct  a  rhombus,  being  given  one  side  and  one  angle. 

Ex.  4.     To  construct  a  rectangle  so  that  two  adjacent  sides  may  be 
equal  to  two  given  lines.     Show  that 
there  is  only  one  solution. 

Ex.  6.  In  an  isosceles  trapezoid  the 
equal  sides  make  equal  angles  with 
each  of  the  other  sides. 

Outline.  Draw  BE  parallel  to  AD. 
Prove  BE,  AD,  BC  all  equal.  Hence, 
prove  angle  ADC  equal  to  BCD. 

Converse  of  theoretn  31, 

156.  Theorem  32.     In  a  quadrangle,  if  the  opposite 
sides  are  equal,  the  figure  is  a  parallelogram. 

[Draw  a  diagonal.     Prove  triangles  equal,  angles  equal,  and  lines 
parallel.  ] 

Ex.     State  and  prove  the  converse  of  the  second  part  of  theorem  31. 

Two  sides  equal  and  parallel, 

157.  Theorem  33.     If  a  quadrangle  has  one  pair  of 
sides  equal  and  parallel,  the  figure  is  a  parallelogram. 

Let  the  quadrangle  ABCD  have  EC  equal  and  parallel  to  AD. 


To  prove  that  ABCD  is  a  parallelogram. 
Draw  a  diagonal  AC. 

Since  the  parallel  lines  BC,  AD  are  met  by  a  transversal 
AC^  the  alternate  angles  BCA,  DAC  are  equal  (117). 


QUAJJRANGLES  79 

Then  in  the  triangles  ABC  and  CDA^  the  side  BC  equals 
AD  (hyp.)  ;  the  side  ^C  is  common ;  and  the  included  angles 
BCA  and  CAD  are  equal ;  therefore  the  triangles  are  equal  (64). 

Thus  the  angles  BAC  and  ACD  (opposite  equal  sides)  are 
equal ;  hence  AB  is  parallel  to  CD  (112). 

Therefore  ABCD  is  a  parallelogram  (def.). 

158.  Cor.  The  lines  joining  the  adjacent  extremities  of  equal 
and  parallel  lines  are  themselves  equal  and  jmrallel. 

Ex.  If  the  parallel  sides  of  a  trapezoid  are  equal,  it  is  a  parallelo- 
gram. 

Diagonals, 

159.  Theorem  34.  The  diagonals  of  a  paraZlelogram 
bisect  each  other. 

A. 


[Compare  two  triangles  formed  by  the  diagonals  and  a  pair  of 
opposite  sides  (as  ABE,  CDE)\  and  thus  prove  the  sides  opposite 
equal  angles  equal.] 

160.  Cor.  I.  The  diagonals  of  a  rhombus  bisect  each  other 
at  right  angles. 

161.  Cor.  2.     The  diagonals  of  a  rhombus  bisect  its  angles. 

Converse  of  theorein  34, 

162.  Theorem  35.  If  the  diagonals  of  a  quadrangle 
bisect  each  other,  the  figure  is  a  parallelogram. 

[Compare  the  two  pairs  of  vertically  opposite  triangles,  and  apply 
64,  112.] 

Ex.  1.  If  the  diagonals  of  a  quadrangle  bisect  each  other  at  right 
angles,  the  figure  is  a  rhombus. 


80  PLANE  GEOMETRY  —  BOOK  I 

Ex.  2.     The  diagonals  of  a  rectangle  are  equal. 

Ex.  3.  Conversely,  if  the  diagonals  of  a  parallelogram  are  equal,  it 
is  a  rectangle. 

Ex.  4.    The  diagonals  of  an  isosceles  trapezoid  are  equal. 

Ex.  5.  Any  line  drawn  through  the  intersection  of  the  diagonals  of 
a  parallelogram  divides  it  into  two  equal  trapezoids.     [Apply  145.] 

Projections, 

163.  Definition.  The  prqjection  of  a  point  upon  a  line 
is  the  foot  of  the  perpendicular  from  the  point  to  the  line. 

The  projection  of  a  line-segment  upon  a  line  is  the  seg- 
ment between  the  projections  of  its  extremities. 


A'        B' 


B 

In  these  figures  the  segment  A^B^  is  the  projection  of  AB. 

164.  Cor.  If  two  lines  are  equal  and  parallel,  then  their 
projections  upon  any  other  line  are  equal. 

[Consider  first  the  case  in  which  the  third  line  is  parallel  to  the  others. 

Next,  when  this  is  not  the  case,  show  that  the  equal  and  parallel 
lines  can  be  made  the  h3rpotenu8es  of  equal  right  triangles  whose 
bases  are  the  projections.] 

Ex.  1.  Parallel  lines  that  have  equal  projections  on  another  line  are 
equal. 

Ex.  2.  Equal  lines  that  have  equal  projections  on  another  line 
make  equal  angles  with  it,  or  else  they  are  parallel  to  it. 

Equal  trapezoids,  or  parallelograms, 

165.  Theorem  36.  Tivo  trapezoids  are  equal  if  they 
have  two  Ojdjacent  sides  and  any  two  opposite  angles  of 
one  equal  to  the  corresponding  parts  in  the  other. 

Outline.  Show  that  the  remaining  angles  are  respectively  equal,  and 
that  the  figures  can  be  superposed  as  in  146. 


QUADRANGLES  81 

166.  Cor.  Two  parallelograms  are  equal  if  they  have  two 
adjacent  sides  and  any  angle  of  one  equal  to  the  corresponding 
parts  in  the  other. 

Ex.  Show  how  to  construct  a  parallelogram,  being  given  two 
adjacent  sides  and  one  angle.  Prove  that  there  is  always  one,  and 
only  one,  solution. 

SERIES    OF    PARALLELS 

167.  Theorem  37.  If  a  series  of  parallel  lines  inter- 
cept  equal  segments  on  any  one  transversal,  then  they 
intercept  equal  segments  on  any  other  transversal. 

Let  the  parallels  make  intercepts  AB,  BC,  CD,  etc.,  on  the 
first  transversal,  and  EF,  FG,  GH,  etc.,  on  the  second  j  and 
let  the  first  set  of  intercepts  be  equal. 


To  prove  that  the  second  set  of  intercepts  are  equal. 

Draw  EK,  FL,  GM,  etc.,  parallel  to  the  first  transversal,  and 
terminated  by  the  successive  parallels. 

Since  EK  and  AB  are  parallels  intercepted  between  paral- 
lels, they  are  equal  (155).  Similarly  FL  equals  BC ;  GiJf  equals 
CD;  etc. 

Now  AB,  BC,  CD,  etc.,  are  all  equal  by  hypothesis,  therefore 
EK,  FL,  GM,  etc.,  are  all  equal. 

Also  in  the  triangles  EKF,  FLG,  GMH,  etc.,  the  angles  KEF, 
LEG,  MGH,  etc.,  are  all  equal  (122) ;  and  the  angles  EFK, 
FGL,  GHM,  etc.,  are  all  equal  (122). 

Therefore  these  triangles  are  all  equal;  and  hence  EFy 
FGy  GHf  etc.,  are  all  equal. 


82  PLANE  GEOMETRY  — BOOK  I 

168.  Definition.  A  series  of  parallels  (as  in  167)  that 
intercept  equal  segments  on  any  transversal  is  called  a 
regular  series  of  parallels. 

169.  Cor.  I.  If  a  regular  series  of  parallels  is  cut  by  two 
transversals,  the  segments  intercepted  on  consecutive  parallels 
have  a  common  difference. 

170.  Cor.  2.  TTie  line  drawn  through  the  m,iddle  point  of 
one  of  the  non-parallel  sides  of  a  trapezoid  parallel  to  the  pair 
of  parallel  sides  bisects  the  remaining  side. 

171.  Cor.  3.  The  line  joining  the  midrpoints  of  the  non- 
parallel  sides  of  a  trapezoid  is  parallel  to  the  other  sides. 

Prove  by  reductio  ad  absurdum,  using  cor.  2  and  116. 

172.  Cor.  4.  T7te  line  joining  the  mid-points  of  the  non- 
parallel  sides  of  a  trapezoid  is  equal  to  half  the  sum  of  the 
X>ardllel  sides. 

The  three  parallels  have  a  common  difference  (169).     Apply  72. 

173.  Cor.  5.  Tlie  line  joining  the  middle  points  of  the  sides 
of  a  triangle  is  parallel  to  the  third  side ;  and  equal  to  its  half 

174.  Cor.  6.  If  one  side  of  a  triangle  is  divided  into  any 
number  of  equal  parts,  and  if  through  the  points  of  division 
parallels  are  draion  to  a  second  side,  then  these  parallels  divide 
the  third  side  into  equal  parts.    (Prove  by  167.) 

175.  Definitions.  A  line  is  said  to  be  trisected  if  it  is 
divided  into  three  equal  parts.  Thus  a  line  has  two  points 
of  trisection. 

One  line  is  said  to  trisect  another  when  the  first  line 
passes  through  one  of  the  points  of  trisection  of  the  other. 

Ex.  1.    Apply  174  to  trisect  a  given  line. 

Outline.  Let  AB  be  the  given  Une.  Draw  another  line  ^A^  at  a 
convenient  angle.  Lay  off  any  three  equal  successive  segments  AL, 
LM,  MN.  Join  NB.  By  114  draw  through  L  and  iW  parallels  to  NB. 
Prove  by  174  that  these  parallels  trisect  the  line  AB. 


QUADRANGLES  83 

Ex.  2.  The  lines  joining  the  mid-points  of  adjacent  sides  of  a  quad- 
rangle form  a  parallelogram  (173). 

Ex.  3.  The  mid-points  of  a  pair  of  opposite  sides  of  a  quadrangle 
and  the  mid-points  of  the  diagonals  are  the  vertices  of  a  parallelogram. 

Medians  of  a  triangle, 

176.  Definition.  In  a  triangle  the  line  joining  any  vertex 
to  the  mid-point  of  the  opposite  side  is  called  a  median 
line  (or  median)  of  the  triangle. 

177.  Theorem  38.  Two  medians  of  a  triangle  tri- 
sect each  other. 


Outline.  In  the  triangle  ABC  let  the  medians  AD  and  BE  inter- 
sect in  0. 

Let  M and  N  be  the  mid-points  of  AO  and  BO.  Show  that  MN  is 
equal  and  parallel  to  DE  (173)  ;  hence  that  MD  is  bisected  at  0,  and 
AD  trisected  at  0. 

Concurrence  of  three  medians, 

178.  Cor.  Tlie  three  medians  of  a  triangle  meet  in  a  com- 
mon point,  and  trisect  each  other. 

179.  Definition.  The  point  of  concurrence  of  the  three 
medians  is  called  the  median  center  of  the  triangle. 

Ex.  Construct  a  triangle  whose  medians  shall  be  equal  to  three 
given  lines. 

Outline  of  analysis.  Prolong  the  median  COF  to  G  so  that  FG 
equals  OF.  Join  AG,  BG.  Prove  AOBG  a  parallelogram.  Show 
that  the  triangle  AOG  has  its  sides  equal  respectively  to  two  thirds  of 
the  medians  ;  and  that  the  problem  is  thus  reduced  to  a  previous  one. 
Give  the  complete  construction,  and  proof. 


84 


PLANE  GEOMETRY — BOOK  I 


Construction  of  Quadrangles 

180.  The  following  problems  are  solved  by  an  extension 
of  the  methods  exemplified  in  131-136.  The  student  should 
make  the  usual  "  preliminary  analysis  "  as  described  in  131, 
and  also  endeavor  to  perform  the  synthetic  construction 
before  consulting  the  solution  given  for  any  problem. 

The  vertices  of  the  quadrangle  will  be  lettered  A,  i?,  C,  Z>, 
in  order ;  and  the  consecutive  sides  ABj  BC,  CD,  DA  will  be 
denoted  by  the  small  letters  a,  6,  c,  d,  in  order. 

Given  two  adjacent  sides  and  three  angles, 

181.  Problem  14.  To  construct  a  quadrangle  hav- 
ing two  a^a/^nt  sides  equal  to  two  given  lines  and 
three  angles  equal  to  three  given  angles,  the  order  in 
which  the  five  parts  are  to  be  taken  being  specified. 

Let  a,  h  be  the  given  lines,  and  A^  C,  2),  the  given  angles. 


L 


b 

\. 

c 

1 

^^--^sik 

CJr-^^^^ 

T 

7 

\ 

The  fourth  angle  B  equals  the  conjunct  of  the  sum  of 
Ay  C,  D  (141),  which  is  obtained  by  subtracting  the  sum  of 
Ay  C,  D  from  a  perigon  as  shown  in  the  figure. 

Lay  off  AB  equal  to  a ;  construct  angle  ABC  equal  to  B ; 
lay  off  BC  equal  to  h.  At  C,  A  draw  lines  CD,  AD  making 
angles  BCD,  BAD  equal  respectively  to  the  given  angles. 

The  quadrangle  ABCD  evidently  fulfills  the  requirements. 

Discussion.  The  limitation  on  the  data  is  that  the  sum  of 
the  three  given  angles  must  be  less  than  a  perigon  (146). 


QUADRANGLES  85 

Given  two  opposite  sides  and  three  angles, 
182.   Problem  15.     To  construct  a  quadrangle  hav- 
ing two  opposite  sides  equal  to  two  given  lines,  and 
three  angles  equal  to  three  given  angles. 

Let  a,  c  be  the  given  lines,  and  A,  B,  c,  the  given  angles. 


D 


/a         b 

C 

-^^ 

1 

a 

-  ^ 

:^ 

Take  AB  equal  to  a,  and  make  angles  ABC  equal  to  B,  BAD 
equal  to  A.  On  BC  take  any  point  P ;  make  angle  BPN  equal 
to  C ;  take  PN  equal  to  c ;  draw  ND  parallel  to  5C,  meeting 
AD  m.  D'^  and  draw  DC  parallel  to  NP,  meeting  BC  in  C. 

The  quadrangle  ABCD  thus  formed  has  the  specified  parts 
equal  to  the  given  parts. 

Prove  that  angle  BCD  equals  C,  and  that  side  CD  equals  c. 

Discussion.  State  limitation  on  data.  Prove  the  solution 
unique  (147). 

Given  three  sides  and  two  included  angles. 

183.  Problem  16.  To  construct  a  quadrangle  hav- 
ing three  sides  equal  to  three  given  lines,  and  the  two 
included  angles  equal  to  two  given  angles. 


I^L 


B  « 

Construct  as  in  134.    State  limitation.    Prove  solution  unique  (146). 


86  PLANE  GEOMETRY — BOOK  I 

Given  three  sides  and  two  angles  adjacent  to  fourth, 

184.  Problem  17.  To  construct  a  quadrangle  hav- 
ing three  sides  equal  to  three  given  lines  and  the  two 
angles  adjacent  to  the  fourth  side  equal  to  two  given 
angles. 

Let  a,  6,  c  be  the  given  lines,  and  A,  D,  the  given  angles. 


Take  AB  equal  to  a ;  make  angle  BAM  equal  A ;  take  any 
point  M  in  the  line  AM,  and  make  the  angle  AMN  equal  D ; 
lay  off  MN  equal  c,  and  draw  NP  parallel  to  AM.  With  B  as 
center  and  radius  equal  to  b,  draw  an  arc  cutting  NP  in  C; 
draw  CD  parallel  to  NM. 

The  quadrangle  ABCb  has  then  the  prescribed  parts. 

Prove  CD  equal  to  c,  and  angle  ADC  equal  to  D. 

Discussion.  In  certain  cases  there  are  two  solutions  to 
the  problem,  viz.  when  the  arc  described  with  B  as  center 
and  radius  equal  to  b  meets  NP  in  two  points  both  situated 
within  the  angle  DAB. 

Show  when  there  is  a  unique  solution,  and  when  none. 


Given  three  sides  and  two  consecutive  angles, 

185.  Problem  18.  To  construct  a  quadrangle  hav- 
ing three  sides  equal  to  three  given  lines  and  two 
consecutive  angles  equal  to  two  given  angles,  one  of 
the  consecutive  angles  being  Oydjacent  to  the  fourth 
side,  and  the  other  not. 


QUADRANGLES  87 

Let  the  parts  a,  c,  d,  A^  B,  be  given. 


Construct  in  succession  DA,  DAB,  AB,  ABM  respectively- 
equal  to  the  given  parts.  With  radius  equal  to  c  and  center 
D,  describe  an  arc  cutting  BM  in  C,  C'. 

State  when  there  is  no  solution,  when  two  solutions,  when  only  one. 

Given  four  sides  and  an  angle, 

186.  Problem  19.  To  construct  a  quadrangle  having 
four  sides  equal  to  four  given  lines  and  one  angle 
equal  to  a  given  angle. 

Let  a,  h,  c,  d  be  the  given  lines,  and  A  the  given  angle. 


1^ 
d 


Let  it  be  required  to  construct  a  quadrangle  whose  sides 
taken  in  order  may  be  equal  to  a,  b,  c,  d,  and  such  that  the 
sides  a  and  d  may  contain  an  angle  equal  to  A. 

Construct  the  triangle  ABD  having  AB  equal  to  a,  AD  equal 
to  d,  and  the  included  angle  DAB  equal  to  A  (133).  Next  on 
BD  construct  the  triangle  BCD  having  BC  equal  to  b,  and  CD 
equal  to  c  (132). 

McM.  ELEM.  GEOM.  —  7 


88  PLANE  GEOMETRY  —  BOOK  I 

Show  by  suitable  figures  that  there  may  be  two  solutions,  only  one 
solution,  or  no  solution. 

Note.  The  case  in  which  three  sides  and  two  opposite  angles  are 
given  is  postponed  (III,  198). 

Ex.  1.  Compare  the  data  in  the  three  unique  solutions  above  with 
the  three  conditions  of  equality  (145-147). 

Ex.  2.  Construct  a  trapezoid  being  given  two  adjacent  sides,  the 
included  angle,  and  the  angle  opposite  to  the  latter.  Which  case  does 
this  come  under  ?    Discuss  the  solution. 


EXERCISES 

1.  Draw  a  line  such  that  its  segment  intercepted  between  two  given 
fixed  indefinite  lines  shall  be  equal  and  parallel  to  a  given  finite  line. 

2.  Draw  a  line  parallel  to  the  base  of  a  triangle,  cutting  the  sides  so 
that  the  sum  of  the  two  segments  adjacent  to  the  base  shall  be  equal 
to  a  given  line. 

Analysis.  Let  PB,  QC  he  the  two  segments.  Draw  PD  parallel 
to  QC.  Then  in  the  triangle  BPD,  the  angles  and  the  sum  of  two 
sides  are  given  (137,  ex.  6).     Give  synthesis. 

3.  Construct  a  parallelogram  being  given  two  adjacent  sides  and  a 
diagonal. 

4.  Construct  a  parallelogram  being  given  a  side  and  two  diagonals. 

5.  Inscribe  a  rhombus  in  a  triangle  having  one  of  its  angles  coinci- 
dent with  an  angle  of  the  triangle. 

6.  One  angle  of  a  parallelogram  is  given  in  position  and  the  point  of 
intersection  of  the  diagonals  is  given  ;  construct  the  parallelogram. 

7.  If  the  diagonals  of  a  quadrangle  bisect  its  angles,  then  it  is  a 
rhombus. 

8.  The  perimeter  of  a  quadrangle  is  greater  than  the  sum  of  its 

diagonals. 

9.  The  sum  of  two  sides  of  a  triangle  is  greater  than  double  the 
median  drawn  to  the  third  side  ;  and  the  perimeter  of  the  triangle 
is  greater  than  the  sum  of  the  three  medians. 

10.  Given  two  medians  and  their  included  angle,  construct  the 
triangle. 


POLYGONS  89 


POLYGONS 


This  section  considers  the  figure  formed  by  any  number 
of  lines  each  of,  which  meets  the  next  in  order,  and  gen- 
eralizes some  of  the  results  obtained  in  the  preceding 
sections. 

187.  Definitions.  A  plane  figure  composed  of  segments  of 
straight  lines  that  inclose  a  portion  of  the  plane  surface,  is 
called  a  polygon. 

These  segments  are  called  the  sides,  their  extremities  the 
vertices,  and  their  sum  the  perimeter,  of  the  polygon. 

Aline  joining  any  two  non-adjacent  vertices  is  called  a 
diagonal. 

The  angles  formed  by  consecutive  sides,  and  situated 
towards  the  interior  of  the  boundary,  are  called  the  interior 
angles  of  the  polygon. 

The  exterior  angles  conjunct  to  these  will  be  called  for 
brevity  the  conjunct  angles. 

If  all  of  the  conjunct  angles  are  convex,  the  polygon  is 
called  a  convex  polygon. 

If  one  of  the  conjunct  angles  is  concave,  the  polygon  is 
said  to  be  concave  at  that  angle. 

In  a  convex  polygon  each  of  the  interior  angles  is  concave, 
and  its  exterior  conjunct  angle  is  convex. 

A  concave  polygon  has  at  least  one  of  the 
conjunct  angles  concave,  and  the  corre- 
sponding interior  angle  convex.  The  sides 
of  this  angle  traverse   the   figure  if  prolonged. 

In  any  polygon  the  concave  angle  formed  by  one  side  and 
the  prolongation  of  an  adjacent  side  is  called  sm  eMerior 
angle  of  the  polygon. 

A  polygon  whose  sides  are  all  equal  is  equilateral,  and 
one  whose  angles  are  all  equal  is  equiangular. 

A  polygon  which  is  both  equilateral  and  equiangular  is 
regular. 


90  PLANE  GEOMETRY  —  BOOK  I 

Two  polygons  that  have  the  sides  of  one  respectively- 
equal  to  the  sides  of  the  oth^,  taken  in  order,  are  said  to  be 
mutually  equilateral^  or  one  is  said  to  be  equilateral  to 
the  other.  ^ 

Two  polygons  that  have  the  angles  of  one  respectively 
equal  to  the  angles  of  the  other,  taken  in  order,  are  said  to 
be  mutually  equiangular^  or  one  is  said  to  be  equiangular 
to  the  other. 

In  two  mutually  equiangular  polygons  the  vertices  of 
equal  angles  are  said  to  correspond;  and  the  sides  join- 
ing corresponding  vertices  are  called  corresponding  sides. 

The  t^o  polygons  are  said  to  be  directly  equiangular  if 
the  sides  of  two  corresponding  angles  can  be  brought  into 
coincidence  in  such  a  way  that  their  corresponding  sides 
may  coincide,  without  turning  either  polygon  out  of  the 
plane ;  otherwise  the  polygons  are  said  to  be  obversely 
equiangular  to  each  other. 

Two  equal  polygons  are  said  to  be  directly  superposahle 
when  they  can  be  superposed  without  turning  either  poly- 
gon out  of  its  plane ;  otherwise  the  equal  polygons  are  said 
to  be  obversely  superposahle. 

The  number  of  interior  angles  in  a  polygon  is  equal  to  the 
number  of  sides. 

A  polygon  of  five  sides  is  called  a  pentagon^  of  six  sides 
a  hexagon  J  of  seven  sides  a  heptagon,  of  eight  sides  an 
octagon,  of  nine  sides  a  nonagon,  of  ten  sides  a  decagon. 
A  twelve-sided  polygon  is  called  a  dodecagon,  and  a 
fifteen-sided  one  a  pentadecagon.  In  the  discussion  of 
general  properties,  a  polygon  of  n  sides  is  called  an 
n-gon. 

The  sum  of  the  n  interior  angles  is  called  the  interior 
angle-sum. 

In  a  convex  polygon,  the  sum  of  the  exterior  angles 
formed  by  prolonging  each  side  one  way,  no  two  adjacent 
sides  being  prolonged  through  the  same  vertex,  is  called 
the  exterior  angle-sum. 


POLYGONS  91 

General  Properties  op  Polygons 

The  following  preliminary  general  theorems  will  be  of 
frequent  use  in  the  theory  of  the  polygon. 

Division  into  triangles  by  diagonals, 

188.  Theorem  39.  In  ^ny  n-gon  if  all  possible  di- 
agonals are  drawn  in  any  manner,  except  that  no  two 
intersect  within  the  polygon,  then  there  will  be  n—3 
such  diagonals,  and  the  n-gon  will  be  divided  into  n— 2 
triangles.* 


Let  the  diagonals  be  drawn  as  stated.  Begin  with  a 
diagonal,  such  as  AC,  that  joins  two  alternate  vertices,  and- 
call  this  the  first  diagonal.  This  first  diagonal  cuts  off  one 
triangle  from  the  n-gon  and  leaves  an  (ji—  l)-gon.  Similarly 
some  second  diagonal  cuts  off  a  second  triangle  from  this 
and  leaves  an  {n  —  2)-gon.  A  third  diagonal  cuts  off  a  third 
triangle  from  the  latter  and  leaves  an  {n  —  3)-gon,  and  so  on. 
When  n—Z  diagonals  are  so  drawn,  there  are  w-— 3  triangles 
cut  off,  and  there  is  left  an  [n-{n-  3)]-gon,  that  is  a  3-gon, 
or  triangle.  Thus  there  are  n  —  3  diagonals  and  w,  —  3  +  1 
triangles.     Hence  the  7Z-gon  is  divided  into  n  —  2  triangles. 

Note.  The  student  who  may  not  be  familiar  with  algebraic  symbols 
may  apply  this  method  of  reasoning  to  the  special  case  of  the  hexagon 
or  heptagon. 

Another  mode  of  proof  consists  in  beginning  with  a  single  triangle, 
and  then  adding  other  triangles,  so  as  to  form  in  succession  a  quad- 
rangle, a  pentagon,  a  hexagon,  etc. 

*  The  symbol  n-2  is  read  n  minus  2,  and  stands  for  the  number 
which  is  2  units  less  than  n. 


92  PLANE  GEOMETRY — BOOK  I 

Interior  angle-sum, 

189.  Theorem  40.  The  sum  of  the  interior  angles  of 
any  n-gon  is  equal  to  n—2  straight  angles. 

[Use  188  and  129.] 

Ex.  1.  An  internal  angle  of  an  equiangular  hexagon  is  equal  to 
twice  the  angle  of  an  equilateral  tridngle  ;  and  that  of  a  regular  oc- 
tagon is  equal  to  a  right  angle  and  a  half. 

Ex.  2.  The  angle  of  a  regular  dodecagon  is  equal  to  five  sixths  of  a 
straight  angle ;  that  is,  equal  to  the  angle  of  a  square  together  with 
the  angle  of  an  equilateral  triangle. 


Exterior  angle-stun, 

190.  Cor.  If  each  side  of  a  convex  n-gon  is  prolonged  one 
way,  no  two  adjcwent  sides  being  extended  through  the  same 
vertex,  then  the  sum  of  the  exterior  angles  so  formed  is  equal  to 
a  perigon. 

For  all  the  exterior  angles  with  all  the  interior  angles 
together  make  up  /i  straight  angles ;  but  the  interior  angles 
alone  make  up  n  —  2  straight  angles ;  hence  the  exterior 
angles  are  together  equal  to  two  straight  angles,  and  there- 
fore equal  to  a  perigon. 

Ex.  1.  Give  an  independent  proof  by  applying  126  as  indicated  in 
figure. 


Ex.  2.     An  exterior  angle  of  an  equiangular  hexagon  is  equal  to 
an  interior  angle  of  an  equilateral  triangle. 

Ex.  3.     The  exterior  angle  of  a  regular  dodecagon  is  equal  to  half 
the  angle  of  an  equilateral  triangle. 


POLYGONS 


93 


EQUALITY   OF   POLYGONS.  —  PRIMARY   CASES 

The  following  four  theorems  relate  to  the  primary  condi- 
tions of  equality  of  two  polygons. 

Mutually  equilateral  and  equiangular, 

191.  Theorem  41.  If  two  polygons  are  muiTiially 
equiangular,  and  have  the  corresponding  sides  equal, 
the  polygons  are  equal. 


[Show  that  the  polygons  are  either  directly  or  obversely  super- 
posable.] 

Note.  In  this  theorem  more  conditions  are  given  than  are  necessary 
to  insure  equality.      This  will  be  evident  from  the  next  two  theorems. 


n 


1  sides  and  n  —  2  included  angles. 

192.  Theorem  42.  //  two  n-gons  have  n—1  sides  of 
one  equal  respectively  ton—1  sides  of  the  other  {taken 
in  order),  and  the  n  —  2  interior  angles  formed  hy  the 
first  set  equal  to  the  corresponding  angles  formed  hy 
the  second  set,  then  the  polygons  are  equal. 


Outline.  Bring  the  equal  parts  into  coincidence  as  in  145,  and  show 
that  this  will  necessitate  the  coincidence  of  the  remaining  side  and  the 
two  remaining  angles  of  one  polygon  with  their  corresponding  parts  in 
the  other. 


94  PLANE  GEOMETRY — BOOK  I   - 

n  —  1  angles  and  n  —  2  intervening  sides, 

193.  Theorem  43.  If  two  n-gons  have  n—  1  angles  of 
one  equal  to  the  corresponding  angles  of  the  other,  and 
the  n  —  2  sides  situated  between  the  vertices  equal  to 
the  corresponding  sides  in  the  other,  then  the  polygons 
are  equal'. 

Outline.  Show  from  the  given  conditions  that  n  —  2  consecutive 
sides  can  be  brought  into  coincidence  with  their  corresponding  parts, 
and  that  the  equality  of  the  angles  necessitates  the  coincidence  of  the 
two  remaining  sides  of  one  polygon  with  the  corresponding  sides  of 
the  other. 

n  —  1  angles  and  n  —  2  consecutive  sides, 

194.  Theorem  44.  If  two  n-gons  have  n—1  angles  of 
one  equal  to  corresponding  angles  of  the  other,  and  any 
n~2  consecutive  sides  of  one  equal  to  the  respective  like- 
placed  sides  of  the  other,  then  the  polygons  are  equal. 

Outline.  Show  that  the  remaining  angle  of  one  is  equal  to  the  re- 
maining angle  of  the  other  ;  and  then  apply  the  preceding  theorem  to 
prove  that  the  polygons  are  equal. 

Construction  of  Polygons 

Problems  20-27  are  concerned  with  the  construction  of  a 
polygon  that  shall  conform  to  certain  prescribed  conditions 
relating  to  the  magnitude  and  position  of  some  or  all  of  its 
parts. 

some  regular  polygons 

Some  of  the  preceding  general  principles  will  be  used  to 
construct  regular  polygons  of  six,  eight,  and  twelve  sides, 
with  the  aid  of  previous  problems.  It  would  be  a  good  ex- 
ercise for  the  student  to  analyze  the  constructions  down  to 
their  simplest  elements,  namely  those  authorized  in  the  con- 
struction postulates.  Some  other  regular  polygons  will  be 
considered  in  Book  III. 


POLYGONS 


96 


195.   Problem   20.     On  a  given  line  to  construct  a 
regular  hexagon. 

Let  AB  be  the  given  line  on  which  a  regular  hexagon  is  to 
be  constructed. 


On  AB  construct  the  equilateral  triangle  AOB-^  continue 
AG,  BO  until  OD  equals  JO,  OE  equals  0B\  join  ED.  Draw 
BC,  DC  parallel  to  OD,  OB ;  and  draw  AF,  EF  parallel  to  OE, 
OA.     The  figure  ABCDEF  is  a  regular  hexagon. 

Outline  proof.  Show  that  each  of  the  triangles  whose  vertices  are 
at  0  is  equal  to  the  triangle  AOB ;  and  hence  that  the  hexagon  is 
equilateral  and  equiangular. 

196.  Problem  21.  On  a  given  line  to  construct  a 
regular  octagon. 

Let  AB  be  the  given  line  on  which  a  regular  octagon  is  to 
be  constructed. 


Since  the  angle  of  a  regular  octagon  is  equal  to  a  right 
angle  and  a  half  (189),  the  figure  may  be  constructed  as 
follows : 

Prolong  AB  to  P,  and  draw  5Jf  perpendicular  to  J5;  draw 
BC  bisecting  the  right  angle  BBM,  and  lay  off  pc  equal  to 


96 


PLANE  GEOMETRY — BOOK  I 


AB.     Similarly  construct  the  side  AH;  draw  CD,  EG  parallel 
to  BM,  and  take  CD  and  HG  each  equal  to  AB.     Draw  G'i'', 
DE,  making  the  angles  NGF,  MDE  each  equal  to  half  a  right 
angle,  and  connect  FE. 
The  figure  ABCDEFGH  is  a  regular  octagon. 

Outline  proof.    Show  that  each  angle  is  equal  to  a  right  angle  and 
a  half.     Also  prove  the  sides  equal. 

197.   Problem  22.     On  a  given  line  to  construct  a 
regular  dodecagon. 

Let  AB  be  the  given  line  on  which  a  regular  dodecagon  is 
to  be  constructed. 

H  O 


Lf- 


rQ 

/     / 

A  JJ 

Since  the  angle  of  a  regular  dodecagon  is  equal  to  the  angle 
of  a  square  plus  the  angle  of  an  equilateral  triangle,  the 
figure  may  be  constructed  as  follows: 

On  the  given  line  AB  construct  a  square  ABPQ.  On  BP 
and  AQ  construct  equilateral  triangles  BPC,  AQN.  On  PC 
and  QN  describe  squares  PCD  V,  QNMR.  On  VD  and  MR  con- 
struct equilateral  triangles  VDE,  MRL ;  and  so  on. 

The  figure  ABCDEFGUKLMN  is  a  regular  dodecagon. 

[Prove  that  the  twelve  sides  are  equal,  and  that  each  angle  is  equal 
to  five  sixths  of  a  straight  angle.] 

Ex.  To  construct  a  pentagon,  being  given  four  angles  and  three 
consecutive  si^es. 


POLYGONS 


97 


TRANSFERENCE   OF    POLYGONS 

198.  To  transfer  a  given  polygon  is  to  construct  another 
polygon  equal  to  the  given  one  so  that  certain  of  its  sides  or 
vertices  may  take  assigned  positions.  This  is  in  accordance 
with  the  postulate  of  figure  transference  (Introd.  12). 

The  first  polygon  is  called  the  trace  of  the  second ;  and 
any  side  or  vertex  of  the  first  is  called  the  trace  of  the  cor- 
responding part  of  the  second. 

General  transference  construction, 

199.  Problem  23.  2'o  transfer  a  given  polygon  so 
that  one  of  the  sides  may  fall  on  a  given  and  equal 
line. 

Let  ABODE  be  the  given  polygon  of  n  sides.  Let  A^B^  be 
equal  to  AB.     It  is  required  to  construct  on  A^B^  a  polygon 


equal  to  ABODE. 


Draw  A'E\  B'o',  making  angles  b'a'e',  a'b'c'  equal,  respec- 
tively, to  BAE,  ABC.  Lay  off  A'e'  equal  to  AE,  and  B'C' 
to  BO.  Continue  this  process  until  n  —  1  sides  and  the  n  —  2 
included  angles  have  been  made  equal  to  their  correspond- 
ing parts;  and  then  complete  the  n-gon  by  joining  the  un- 
connected ends  of  the  last  two  segments. 

The  two  polygons  are  equal  (192). 

Note.  If  A'E'  and  B'C  be  drawn  at  the  other  side  of  A'B'^  and 
the  polygon  be  completed  as  before,  the  new  polygon  is  obversely 
superposable  on  the  given  one  (187). 


98 


PLANE  GEOMETRY — BOOK  I 


Translation  construction, 

200.  Problem  24.  To  transfer  a  given  polygon  so 
that  corresponding  sides  in  the  two  positions  shall  be 
parallel,  and  so  that  the  lines  joining  corresponding 
vertices  may  be  equal  and  parallel  to  one  and  the  same 
given  line. 

Let  ABCB  be  the  given  polygon,  and  L  the  given  line. 


To  construct  a  polygon  a'b'c'd'  equal  to  ABCD,  so  that 
the  sides  a'b',  B'c'f  C'D\  B'A'  may  be  respectively  equal  and 
parallel  to  AB,  BC,  CD,  DA,  and  so  that  AA\  BB\  CC\  DD^  may 
be  each  equal  and  parallel  to  L. 

Draw  AA\  bb\  CC\  DD'  each  equal  and  parallel  to  L.  Join 
A'B',  B'C',  C'D',  D'A'. 

Then  A'B'CfD'  fulfills  the  given  conditions. 

[Prove  corresponding  sides  of  the  polygons  equal  and  parallel  and 
corresponding  angles  equal.  ] 

Note.  This  kind  of  transference  is  called  translation.  The  given 
line  is  called  the  line  of  translation.  The  construction  used  in  this 
problem  is  called  the  translation  construction. 

201.  Definition.  Two  equal  polygons  are  said  to  be  siw^i- 
larly  placed  when  any  side  and  its  corresponding  side  are 
parallel  and  are  on  the  same  side  of  the  line  joining  corre- 
sponding extremities. 

Thus  if  a  polygon  is  translated  as  in  200,  the  polygon  and 
its  trace  are  similarly  placed. 


POLYGONS  99 

The  rotation  construction, 

202.  Problem  25.  To  transfer  a  given  polygon  so 
that  one  verte^v  may  he  unchanged,  and  so  that  each 
side  may  make  with  its  trace  an  angle  equal  to  one 
and  the  same  given  angle. 

Let  ABCD  be  the  given  polygon,  A  the  vertex  that  is  to  be 
unchanged,  and  L  the  given  angle. 


To  construct  an  equal  polygon  AB'c'd'  so  that  the  angle 
between  corresponding  sides  (or  their  prolongation)  shall  be 
equal  to  L. 

Join  A  to  the  other  vertices.  Turn  the  lines  AB,  AC,  AD 
in  the  same  sense  through  an  angle  equal  to  L,  into  the  posi- 
tions AB',  AC',  ad'.  In  other  words,  make  the  angles  BAB', 
CAC',  DAD'  each  equal  to  L,  and  make  AB'  equal  to  AB,  AC* 
to  AC,  AD'  to  AD.     Join  B'c',  C'D'. 

Since  the  angles  BAB'  and  CAC'  are  equal,  the  angles  BAG 
and  b'AC'  are  equal ;  hence  the  triangles  BAC  and  B'AC'  are 
equal. 

Similarly  the  triangles  CAD  and  C'AD'  are  equal. 

Hence  the  polygons  ABCD  and  AB'C'd'  are  equal. 

Next,  to  prove  that  the  angle  formed  by  the  prolongations 
of  the  sides  BC  and  b'C'  equals  the  angle  L. 

Let  BC  meet  AC'  in  P;  and  let  BC  prolonged  meet  B'C*  in 
0  (not  shown  in  figure). 

The  angles  ACB  and  AC'b'  have  been  proved  equal;  hence 
the  triangles  FAC  and  POC'  have  two  angles  of  one  equal  to 


100  PLANE  GEOMETRY  —  BOOK  I 

two  angles  of  the  other ;  therefore,  the  third  angles  PAC  and 
POC'  are  equal. 

But  the  angle  PAC  equals  i;  therefore  POC'  equals  the 
given  angle. 

Similarly  the  angle  formed  by  CD  and  C'i)'  equals  the 
given  angle. 

203.  Definition.  The  kind  of  transference  described  in 
202  is  called  rotation.  The  fixed  vertex  is  called  the  cen- 
ter of  rotation;  and  the  given  angle  the  angle  of  rotation. 
The  construction  used  is  called  the  rotation  construction. 

204.  Again,  if  any  point  O  is  taken  in  the  plane  of  the 
polygon  and  joined  with  the  vertices,  and  if  the  whole  figure 
is  turned  about  o  by  a  similar  construction,  then  0  is  called 
the  center  of  rotation. 

205.  Construction  for  center  of  rotation.  It  will  next  be 
shown  how  to  find  a  center  of  rotation  by  means  of  which  a 
given  polygon  can  be  transferred  to  any  other  given  position. 
We  begin  with  the  simpler  problem  of  rotating  a  line. 

Center  of  rotation  for  transferring  a  line. 

206.  Problem  2^.  Given  two  equal  lines  not  -paral- 
lel, and  not  in  tlie  same  straight  line ;  to  find  a  point 
in  the  plane  such  that  it  may  he  taken  as  a  center  of 
rotation  for  the  purpose  of  transferring  one  line  into 
coincidence  with  the  otJier. 

Let  AB  and  a'b'  be  the  two  equal  ^B' 

lines.  ^'^^  /  \ 

.^^'        /    • 

It   is   required  to  find  a  point  0,     q^:"^'  _  _    /     J 

such  that  if  it  be  joined  to  the  ex-  jVnT^*       /           J 

tremities   of  the  lines,  the  triangle  ;  \  \  /^\        j 

OAB  can  be  rotated  about  0  into  the  '>-''^l— — --^'r 

position  OA!b\  ^ 


POLYGONS  101 

Analysis.     Suppose  the  point  0  satisfies  the  condition. 
Then  GAB  is  directly  superposable  on  OA^B\ 
Therefore  OA  is  equal  to  0A\  and  OB  to  0B\ 
Hence,  0  lies  on  the  line  bisecting  AA^  at  right  angles ; 

and  also  on  the  line  bisecting  BB'  at  right  angles.     It  is 

therefore  to  be  determined  as  follows  : 

Construction.  Join  AA\  bb'.  Draw  the  perpendicular  bi- 
sector of  each  of  these  lines.  Let  the  perpendiculars  meet 
in  0.     Then  O  is  the  required  center  of  rotation. 

Prove  by  showing  that  OAB  is  directly  superposable  on  OA'ff. 

Discussion.  There  is  another  solution  if  A  is  taken  to 
correspond  to  B',  and  B  to  A'. 

If  the  lines  AB  and  A'b'  are  parallel  as  well  as  equal,  and 
if  they  are  on  the  same  side  of  the  line  AA',  the  two  lines 
bisecting  AA',  BB'  perpendicularly  do  not  meet,  and  there  is 
no  center  of  rotation.  In  this  case  the  line  AB  can  be  trans- 
ferred to  the  position  A'b'  by  translation.  Show  that  there 
is,  however,  a  center  of  rotation  that  transfers  B  to  the 
position  A',  and  A  to  the  position  B'. 

Consider  the  special  case  in  which  the  mid-points  of  AA' 
and  BB'  coincide  at  a  point  M.  Show  that  M  is  then  the 
required  center. 

Center  of  rotation  for  transferring  a  polygon. 

207.  Problem  27.  Given  two  equal  and  directly 
superposable  polygons,  ivhich  are  not  similarly  placed, 
to  find  a  center  of  rotation  in  order  to  transfer  one 
polygon  into  coincidence  with  the  other. 

Outline.  By  the  last  problem,  find  a  point  0  about  which  one  side 
AB  may  be  rotated  to  coincidence  with  A'B',  in  which  A  corresponds 
to  A',  and  B  to  B'.     Show  that  the  polygons  will  then  coincide. 

Note.  Of  two  equal  and  similarly  placed  polygons,  one  can  be 
transferred  to  the  position  of  the  other  by  translation  (200). 

Ex.  Show  how  to  rotate  a  given  square  into  coincidence  with  any 
given  equal  square.     Show  that  in  this  case  there  are  four  solutions. 


102  PLANE  GEOMETRY— BOOK  I 

Rotation  through  a  straight  angle, 

208.  Theorem  45.  //  a  polygon  is  rotated  through 
a  straight  angle  about  any  point  of  its  plane,  then 
any  side  and  its  trace  are  parallel,  hut  lie  on  opposite 
sides  of  the  line  joining  corresponding  extremities. 

Let  ABCD  be  the  polygon,  O  the  center  of  rotation.  Let 
the  lines  OAy  OB,  00,  OD  be  each  turned  through  a  straight 
angle  into  the  opposite  position  0A\  OB',  OC',  OB', 

C 


To  prove  that  any  side  AB  and  its  trace  a'b'  are  parallel, 
and  on  opposite  sides  of  the  line  AA'. 
[Compare  the  triangles  O^B  and  OA'B'."] 

209.  Definition.  Two  equal  polygons  are  said  to  be  oppo- 
sitely placed  when  any  side  and  its  corresponding  side  are 
parallel  and  are  on  opposite  sides  of  the  line  joining  corre- 
sponding extremities.     (See  figure  in  208.) 

210.  Cor.  I.  If  a  polygon  is  turned  through  a  straight 
angky  the  polygon  and  its  tra^e  are  oppositely  placed. 

211.  Cor.  2.  If  two  polygons  are  equal  and  oppositely  placed, 
the  lines  joining  corresponding  points  meet  in  the  same  point 
and  bisect  each  other. 

[The  diagonals  of  a  parallelogram  bisect  each  other.] 

Center  of  Rotation. 

211  (a).  Cor.  3.  If  tivo  polygons  are  equal  and  oppositely 
placed,  one  of  them  can  he  transferred  into  the  position  of  the 
other  by  rotation  through  a  straight  angle  about  the  intersection 
of  the  lines  joining  corresponding  vertices. 

Prove  by  means  of  211.  Also  show  that  this  is  the  "  special  case  " 
referred  to  in  the  discussion  of  problem  26. 


POLYGONS  103 

Equiangular  polygons  placed  in  parallelism,, 
212.  Theorem  46.  //  any  two  polygons  are  directly 
equiangular,  and  if  they  are  placed  so  as  to  have  a 
pair  of  corresponding  sides  parallel,  then  each  side  of 
one  is  parallel  to  the  corresponding  side  of  the  other. 
Let  the  equiangular  polygons  ABCD,  A'b'c'D^  have  the 
corresponding  angles  A  and  ^',  B  and  B',  c  and  C\  D  and  D\ 
occurring  in  direct  order.      Let  the  sides  AB  and  A'b'  be 

parallel. 

(J 
To  prove  that  BC  is  paral-  /\^ 

lei  to  5'C',  CD  to  C'D',  DA  to  /gr_\v  - C 

First,  let  the  parallels  AB         \       ^y^r \-—-^B' 

and  A^B^  lie  on  the  same  side  \^ \^ 

A  A 

of  the  line  AA}. 

By  the  translation  construction  (prob.  24)  transfer  the 
polygon  A'B'C'D'  (using  A  A'  as  the  line  of  translation)  so 
that  A'  may  coincide  with  A. 

Since  the  sides  are  equal  and  parallel  to  their  traces,  A'b' 
falls  on  the  parallel  line  AB ;  let  it  take  the  position  AB". 
Since  the  polygons  are  directly  equiangular,  A'D'  falls  on  AD ; 
let  it  take  the  position  AD". 

Since  the  angle  D"  is  equal  to  D,  and  B"  to  B,  it  follows 
that  the  sides  of  ABCD  are  parallel  to  those  of  ab"C"d"  and 
therefore  to  those  of  A'b'c'd'. 

Next  let  the  parallels  AB  and  a'b'  lie  on  opposite  sides  of 
the  line  AA'.  ^  ,, 

Kotate  the  polygon  A'b'C'd'  A^>..^^  /    ^--^ 

through  a  straight  angle  about        /  ^^^    -OV^^  4  /  ^ 

a  point  in  its  plane  into  the      /  y  ^.^-^^ 

position  ^1^1  CiDi  (208).  ,^\         /     """^'^b'^      ^ 

The  side  A^B^  is  parallel  to  \.<^'  ^""^^ 

AB,  and  on  the  same  side  of 

the  line  AA^;  hence,  by  the  first  part,  the  sides  of  ABCD  are 
parallel  to  those  of  A^B.C^D^  and  therefore  to  those  of  a'b'&D'. 

MOM.   ELEM.   GEOM.  —  8 


104  PLANE  GEOMETRY — BOOK  I 

213.  Cor.  If  two  polygons  are  directly  equal  and  have  a  pair 
of  corresponding  sides  parallel,  the  polygons  are  either  similarly 
placed  or  oppositely  placed. 

Axial  Symmetry 

The  theory  of  symmetric  figures  is  of  importance  in  con- 
nection with  the  equality  of  polygons.  We  begin  with  the 
case  of  two  symmetric  points. 

214.  Definition.  Two  points  are  said  to  be  symmetric 
to  each  other  with  regard  to  a  certain  line,  called  the  axis 
of  symmetry,  if  the  line  joining  the  two  points  is  bisected 
perpendicularly  by  the  axis. 

Pritnary  construction, 

216.   Problem  28.     To  construct  tJie  symmstric  point 
of  a  given  point  with  regard  to  a  given  axis. 
Let  AA^  be  the  given  axis,  and  P  the  given  point. 

»F 
4 10 Af' 

\p' 

Draw  PO  perpendicular  to  AA^  and  prolong  it  to  J*'  so  that 
OP'  equals  PO. 

The  points  P  and  P'  are  symmetric,  by  definition. 

Discussion.  Prove  that  a  point  has  only  one  symmetrical 
point  with  regard  to  a  given  axis. 

Show  that  the  construction  does  not  apply  if  the  given 
point  lies  on  the  given  axis.  Show  that  if  P  be  taken  nearer 
and  nearer  to  the  axis,  then  P'  comes  nearer  and  nearer  to  P. 

This  fact  suggests  the  definition  that  follows. 

216.  Definition.  Any  point  on  the  axis  will  be  said  to 
have  its  symmetric  point  coincident  with  itself. 

217.  Definition.  The  construction  just  given  is  called 
reflection.  To  reflect  a  given  point  with  regard  to  a  given 
axis  is  to  find  its  symmetric  point. 


POLYGONS  105 

218.  Definition.  Any  two  lines  (straight  or  curved)  are 
said  to  be  syrmnetric  lines  with  regard  to  a  given  axis 
when  every  point  of  each  line  has  its  symmetric  point  on 
the  other  line. 


219.  Definition.  Any  two  figures  are  said  to  be  sym- 
metric figures  with  regard  to  an  axis  when  their  bounding 
lines  are  symmetric. 

Superposition  by  folding  over, 

220.  Theorem.  47.  If  the  portion  of  the  plane  at 
one  side  of  the  axis  is  conceived  to  he  revolved  about 
the  axis  {or  folded  over),  so  that  it  coincides  with  the 
portion  at  the  other  side  of  the  axis,  then  every  point 
of  the  plane  will  coincide  with  its  symmetric  point. 

Let  P,  P'  be  any  two  symmetric  points  with  regard  to  ^^l'. 

,P 


To  prove  that  P  and  P'  may  be  made  to  coincide  by  re- 
volving the  portion  of  the  plane  on  one  side  of  AA^  into 
coincidence  with  the  other  portion. 

Let  the  line  PP'  meet  AA'  at  0. 

By  definition  PP^  is  bisected  at  right  angles  by  AA\ 

The  revolution  of  either  portion  of  the  plane  about  OA 
leaves  all  points  of  OA  unchanged,  and  the  angles  AOP,  AOP' 
remain  right  angles. 

Hence  the  lines  OP  and  OP'  come  into  coincidence,  other- 
wise the  right  angles  would  not  be  equal. 

Then,  since  OP  equals  OP',  the  points  P  and  P'  come  into 
coincidence. 


106 


PLANE  GEOMETRY — BOOK  I 


221.  Cor.  I.  Any  Jigure  can  be  brought  into  coincidence 
with  its  symmetric  jigure  (or  reflection)  by  folding  over. 

222.  Cor.  2.  If  two  points  are  symmetiric  to  two  other  points 
respectively,  the  line-segment  joining  the  first  two  is  symmetric 
loith  the  line-segment  joining  the  other  two. 

223.  Cor.  3.  Tlie  symmetric  figure  of  a  finite  straight  line 
is  an  equal  straight  line. 

224.  Cor.  4.  Tlie  symmetric  figure  of  an  indefinite  straight 
line  is  another  such  line,  and  the  two  lines  m^ke  equal  angles 
with  the  a^is  of  symmetry. 

225.  Cor.  5.  Tlie  symmetric  figure  of  a  2)lane  angle  is  an 
equal  plane  angle. 

Symmetric  polygons, 

226.  Problem  29.  To  construct  the  figure  symmetric 
to  a  given  polygon  with  regard  to  a  given  axis  of  sym- 
metry. 

Let  ABCD  be  the  given  polygon,  LL'  the  given  axis  of 
symmetry. 

To  construct  the  figure  symmetric 
to  ABCD  with  regard  to  LL\ 

Construct  the  symmetric  points 
of  the  vertices  with  regard  to  XL' 
(215). 

Join  the  new  points  in  the  same 
order  as  their  symmetric  points  are 
joined. 

The  polygon  A'b'c'd'  so  formed  is  symmetric  to  ABCD. 

Prove  by  222  and  definition  in  219. 

227.  Definition.  The  figure  symmetric  to  a  given  figure 
is  called  its  reflection  or  image  with  regard  to  the  given 
axis.  As  the  two  figures  are  obversely  superposable,  this 
construction  will  be  called  the  ohversion  construction. 


POLYGONS 


107 


228.  Cor.  I.  If  tivo  polygoiis  are  obversely  equals  it  is  pos- 
sible to  translate  one  of  them  so  as  to  he  symmetric  to  the  other 
with  regard  to  some  axis. 

229.  Cor.  2.  If  two  polygons  are  obversely  equal,  it  is  pos- 
sible to  rotate  one  so  as  to  be  symmetric  to  the  other  vnth  regard 
to  a  given  axis. 

AXIS    OF    SYMMETRY    OF    A    FIGURE 

230.  Definition.  If  a  straight  line  divides  a  figure  into 
two  parts  that  are  symmetric  with  respect  to  that  line  as  an 
axis,  the  figure  is  said  to  be  a  symmetric  figure,  and  the 
line  is  called  an  axis  of  symmetry  of  the  figure. 

E.g.  in  an  isosceles  triangle  the  bisector  of  the  vertical 
angle  is  an  axis  of  symmetry. 

Some  figures  have  two  or  more  axes  of  symmetry.  A 
rectangle  has  two  axes  of  symmetry,  a  rhombus  two,  and  an 
equilateral  triangle  three  ;  hence  the  rectangle  and  rhombus 
are  said  to  have  biaxial  symmetry;  and  the  equilateral 
trangle  to  have  triaxial  symmetry. 

Ex.  1.  A  square  has  four  axes  of  symmetry,  a  regular  pentagon 
five,  and  a  regular  hexagon  six. 


W-. — 


Ex.  2.  If  two  points  are  symmetric  as  to  an  axis,  and  if  each  of 
them  is  reflected  with  regard  to  another  axis  perpendicular  to  the  first, 
then  the  two  points  and  their  two  reflections  are  at  the  vertices  of  a 
rectangle  ;  and  this  rectangle  is  a  symmetric  figure  with  regard  to  each 
of  the  two  given  axes. 


108  PLANE  GEOMETRY — BOOK  I 

Central  Symmetry 

231.  Definition.  Two  points  are  said  to  be  symmetric  with 
regard  to  a  fixed  point,  called  the  center  of  symmetry,  when 
the  line  joining  the  two  points  is  bisected  at  the  center. 

The  line  joining  any  point  to  the  center  of  symmetry  is 
called  its  radius  of  symmetry. 

Prhnary  constt^uction, 

232.  Problem  30.  To  construct  the  symmetric  point 
of  a  given  point  with  regard  to  a  given  center. 

Let  0  be  the  given  center,  and  P  the  given  point. 

P' 0_ p 


Draw  the  radius  of  symmetry  PO  and  prolong  it  so  that 
OP'  equals  PO. 

The  points  P  and  P'  are  symmetric  by  definition. 

Discussion.  Prove  that  a  given  point  has  only  one  sym- 
metric point  with  regard  to  a  given  center. 

Show  that  if  P  be  taken  nearer  and  nearer  to  O,  then  P' 
comes  nearer  and  nearer  to  O.  This  fact  suggests  the  defini- 
tion that  follows: 

233.  Definition.  The  center  of  symmetry  will  be  said  to 
have  its  symmetric  point  coincident  with  itself. 

234.  Cor.  The  point  P  can  be  brought  into  coincidence  with 
its  symmetric  point  P'  by  revolving  the  radius  of  symmetry  OP 
through  two  right  angles  into  the  position  OP'. 

235.  Definition.  Any  two  lines  (straight  or  curved)  are 
said  to  be  symmetric  lines  with  regard  to  a  given  center 
when  every  point  of  each  has  its  symmetric  point  on  the 
other  line. 


POLYGONS  109 

236.  Definition.  Any  two  figures  are  said  to  be  ST/m- 
metric  figures  with  regard  to  a  given  center  when  their 
bounding  lines  are  symmetric. 

Symmetric  line-segments, 

237.  Theorem  48.  //  two  points  are  syrrvmetric  to 
two  other  points  respectively,  the  line  joining  the  first 
two  is  symmetric  to  the  line  joining  the  other  two. 

Let  0  be  the  center  of  symmetry ;  and  let  P  be  symmetric 
to  P',  and  Q  to  Q'. 


To  prove  that  the  line  PQ  is  symmetric  to  P*Q\ 

Take  any  point  i?  in  PQ.  Draw  RO  and  prolong  it  to  meet 
P'Q'  in  R\ 

Rotate  the  triangle  OPQ  about  O  through  a  straight  angle 
so  that  OP  falls  on  OP' ;  then,  by  the  equality  of  angles  and 
sides,  OQ  falls  on  OQ' ;  hence  PQ  falls  on  P'Q\  But  OR  falls 
along  OR'  by  the  equality  of  angles.  Therefore,  the  point  R 
falls  on  R'  ;  hence  OR'  equals  OR,  and  the  points  R  and  i?'  are 
symmetric. 

Therefore,  any  point  in  PQ  has  its  symmetric  point  in 
P'Q\   and  thus   the   lines   PQ  and   P'Q'   are  symmetric. 

238.  Cor.  I.  The  symmetric  Jigure  of  a  Jinite  straight  line 
is  an  equal  and  parallel  line. 

239.  Cor.  2.  The  symmetric  Jigure  of  a  triangle  is  another 
triangle  equal  and  oppositely  placed. 

240.  Cor.  3.  The  symmetric  Jigure  of  a  plane  angle  is  an 
equal  angle,  whose  sides  are  parallel  to  the  sides  of  the  first. 


110  PLANE  GEOMETRY — BOOK  I 

Symmetric  polygons, 

241.   Problem  31.     To  construct  the  symmetric  figure 
of  a  given  polygon  with  regard  to  a  given  center. 
Let  ABCD  be  the  given  polygon,  O  the  center  of  symmetry. 


To  construct  the  symmetric  figure  of  ABCD. 

Find  the  points  A\  B',  C',  D',  symmetric  respectively  to 
the  vertices  A,  By  C,  D. 

Join  these  points  in  the  same  order  as  the  given  vertices 
are  joined.    Then  A'b'c'd'  is  symmetric  to  ABCD.    [236,  237.] 

Note.  The  two  polygons  are  directly  equal,  but  oppositely  placed. 
They  can  be  brought  into  coincidence  by  rotating  either  of  them  through 
a  straight  angle  about  O. 

242.  Cor.  I.  If  tivojyolygons  are  directly  equal,  it  is  possible 
to  rotate  one  of  them  so  as  to  be  symmetric  to  the  other  with 
regard  to  a  given  center  of  symmetry. 

243.  Cor.  2.  Any  two  equal  and  oppositely  placed  polygons 
have  a  center  of  symmetry. 

244.  Definition.  A  single  figure  is  called  a  sym,m^tri^ 
figure  as  to  a  certain  center,  if  any  point  on  the  boundary 
has  its  symmetric  point  also  on  the  boundary. 

E.g.  a  parallelogram  is  symmetric  as  to  the  intersection 
of  its  diagonals. 

Ex.  1.    An  equilateral  triangle  has  no  center  of  symmetry. 
Ex.  2.     A  regular  hexagon  has  a  center  of  symmetry. 
Ex.  3.     No  pentagon  has  a  center  of  symmetry. 
Ex.  4.     Any  polygon  of  an  even  number  of  sides  whose  opposite 
sides  are  equal  and  parallel  has  a  center  of  symmetry. 


POLYGONS 


111 


jBi-axial  related  to  central  symmetry, 

245.  Theorem  49.  Two  points  that  are  symmetric 
to  the  same  point  with  regard  to  two  rectangular  axes, 
respectively,  are  symmetric  to  each  other  as  to  the 
intersection  of  these  axes. 

Let  OL,  DM  be  two  axes  at  right  angles.  Let  the  symmet- 
ric points  of  P  as  to  these  axes  be  Q,  R,  respectively. 


-iP 


To  prove  that  Q  and  R  are  symmetric  as  to  0. 

Join  OQ,  OR.  Prove  that  the  triangles  RLO  and  OMQ  are 
equal ;  that  the  angles  ROL,  LOM,  MOQ  are  together  equal  to 
a  straight  angle ;  and  that  ROQ  is  a.  straight  line. 

246.  Cor.  I.  Two  figures  that  are  symmetric  to  the  same 
figure  with  regard  to  two  rectangular  axes,  respectively,  are 
symmetric  to  each  other  as  to  the  intersection  of  these  axes. 

.247.   Cor.  2.     If  a  single  figure  has  two  axes  of  symmetry 
at  right  angles,  their  intersection  is  a  center  of  symmetry. 


ea 


Ex.  If  a  figure  has  a  center  of  symmetry  and  an  axis  of  symmetry, 
then  the  center  lies  on  the  axis ;  and  there  is  another  axis  of  symmetry 
perpendicular  to  the  first. 


112  PLANE  GEOMETRY— BOOK  I 


LOCUS  PROBLEMS 

248.  Many  geometric  problems  are  concerned  with  finding 
the  position  of  a  point  so  that  it  may  fulfill  certain  prescribed 
conditions. 

It  may  happen,  however,  that  the  prescribed  conditions 
do  not  suffice  to  fix  the  point  entirely,  but  are  sufficient  to 
restrict  it  to  some  line  or  group  of  lines.  Hence  arises  the 
idea  of  a  locuSj  which  may  be  defined  as  follows. 

249.  Definition.  If  every  point  on  a  certain  line  or  group 
of  lines  (straight  or  curved)  satisfies  prescribed  conditions, 
and  if  no  other  point  does  so,  then  that  line  or  group  of 
lines  is  called  the  locios  of  the  points  fulfilling  those  con- 
ditions. 

The  locus  may  be  conveniently  imagined  as  the  path 
traced  by  a  moving  point  that  continues  to  satisfy  the  pre- 
scribed conditions. 

250.  Method  of  investigation.  The  investigation  of  a  locus 
may  be  divided  into  an  analysis  and  a  synthesis. 

Analysis.  Take  any  point  and  suppose  it  to  satisfy  the 
prescribed  conditions.  By  changing  its  position  subject  to 
these  conditions,  try  to  form  some  idea  of  the  path  of  the 
moving  point,  noting  any  special  positions  which  it  passes 
tlirough,  and  thus  endeavor  to  discover  what  fixed  line 
(straight  or  curved)  is  traced  by  the  moving  point. 

Draw  such  line  and  try  to  demonstrate,  by  means  of  the 
given  conditions,  that  it  is  actually  fixed.  When  this  is 
done,  the  analysis  is  completed. 

Synthesis.  Draw  the  logical  inference  from  the  preceding 
that  every  point  satisfying  the  conditions  must  lie  on  the 
fixed  line  in  question. 

The  next  step  is  to  prove  conversely  that  every  point  on 
the  line  satisfies  the  prescribed  conditions. 


LOCUS  PROBLEMS 


113 


It  can  then  be  concluded  that  this  line  contains  all  those 
points  (and  those  only)  which  fulfill  the  conditions  assigned. 

Note.  The  word  line  is  here  used  as  an  abbreviation  for  the  phrase 
"  line  or  group  of  lines  (straight  or  curved)." 

251.  This  method  may  be  briefly  summarized  thus : 

In  order  to  prove  that  a  certain  line  L  is  the  locus  of  a 
point  P  fulfilling  the  condition  A,  it  is  necessary  and  suf- 
ficient to  demonstrate  the  following  two  converse  proposi- 
tions : 

(1)  If  P  fulfills  the  condition  A,  then  P  lies  on  the  line  i; 

(2)  If  P  lies  on  z,  then  P  satisfies  A. 

Instead  of  proving  (1)  it  may  sometimes  be  more  con- 
venient to  prove  its  contraposite : 

If  P  does  not  lie  on  L,  then  P  does  not  satisfy  A.  Show 
that  this  proposition  is  equivalent  to  (1).     See  Arts.  105, 106. 

252.  Definition.  For  convenience  the  line-segment  con- 
necting two  points  is  sometimes  called  the  Join  of  one  point 
to  the  other. 

253.  Problem  32.  To  find  the  locus  of  a  point  such 
that  its  joins  to  two  given  points  are  equal. 

Let  A  and  B  be  the  two  given  points.  Let  P  be  any  point 
such  that  the  lines  PA  and  PB  are  equal. 


/ 


\ 


M 


B 


To  find  and  construct  the  locus  of  P. 


114  PLANE  GEOMETRY — BOOK  I 

Analysis.  Taking  successive  positions  of  P  subject  to  the 
condition  that  PA  equals  PB,  and  observing  that  M  the  mid- 
point oi  AB  fulfills  this  condition,  we  are  led  to  surmise  that 
the  path  of  P  is  the  straight  line  joining  P  to  the  mid-point 
of  AB. 

To  prove  that  this  line  PM  remains  actually  fixed  as  P 
moves. 

The  triangles  PAM  and  PBM  have  their  sides  respectively 
equal.  Therefore  the  angles  PMA  and  PMB  are  equal,  and 
each  equal  to  a  right  angle. 

Hence  MP  is  perpendicular  to  the  fixed  line  ^5  at  its  mid- 
dle point,  and  is  therefore  fixed,  no  matter  what  position 
the  moving  point  P  takes  while  subject  to  the  assigned  con- 
dition. 

It  follows  that  P  moves  along  this  fixed  line. 

Synthesis.  It  has  now  been  proved  that  every  point  which 
satisfies  the  prescribed  condition  lies  on  the  perpendicular 
bisector  MP. 

It  remains  to  be  proved  conversely  that  every  point  on  MP 
satisfies  the  condition. 

-    Let  P  be  any  point  on  the  perpendicular  bisector.     Join 
PA  and  PB. 

The  two  right  triangles  have  the  sides  about  the  right 
angles  respectively  equal ;  therefore  the  hypotenuses  PA 
and  PB  are  equal. 

Hence  all  those  points  (and  those  only)  that  satisfy  the 
given  condition  lie  on  the  line  MP. 

Therefore  the  perpendicular  bisector  extended  indefinitely 
both  ways  is  the  locus  required. 

Ex.  1.  Find  the  locus  of  the  vertex  of  an  isosceles  triangle  whose 
base  is  given  in  magnitude  and  position. 

Ex.  2.  Show  that  a  simple  construction  for  bisecting  a  given  line- 
segment  can  be  derived  from  the  locus  problem  solved  above.  (Com- 
pare 70.) 

Ex.  3.  Show  that  this  locus  problem  also  furnishes  a  solution  to  the 
problem  of  erecting  or  dropping  a  perpendicular  to  a  given  line. 


LOCUS  PROBLEMS 


115 


254.  Problem  33.  To  find  the  locus  of  a  point  from 
which  the  perpendiculars  to  two  given  intersecting  lines 
are  equal. 

Let  AB  and  CD  be  the  two  given  lines.  Let  P  be  any 
point  such  that  PM,  the  perpendicular  to  AB,  is  equal  to  PN, 
the  perpendicular  to  CD. 


To  find  the  locus  of  P. 

Analysis.  Taking  successive  positions  of  P  subject  to  the 
given  condition,  and  observing  that  the  moving  point  can 
come  nearer  and  nearer  to  0,  we  are  led  to  think  that  0  is  a 
point  on  the  locus,  and  that  the  straight  line  OP  is  the  path 
of  P. 

To  prove  that  this  line  is  actually  fixed  in  position  as  P 
moves. 

The  triangles  POM,  PON  have  two  sides  of  one  respectively 
equal  to  two  sides  of  the  other,  and  the  angles  opposite  a 
pair  of  equal  sides  are  right  angles ;  therefore  the  angles 
POM  and  PON  are  equal  (98). 

HeuQe  OP  is  a  bisector  of  the  angle  HON,  and  is  therefore 
a  fixed  line. 

It  follows  that  the  point  P  is  on  one  or  other  of  the  bi- 
sectors of  the  angles  contained  by  the  two  given  intersecting 
lines. 

Synthesis.  It  has  now  been  proved  that  every  point  which 
satisfies  the  prescribed  condition  lies  on  one  or  other  of  the 
two  angle-bisectors. 

It  will  next  be  proved  conversely  that  every  point  on 
either  of  these  lines  satisfies  the  condition. 


116  PLANE  GEOMETRY  —  BOOK  I 

Let  Q  be  any  point  on  either  angle-bisector.  Draw  QR 
perpendicular  to  AB^  and  QS  perpendicular  to  CD. 

The  two  right  triangles  QOR,  QOS  have  the  acute  angles 
QOR  and  QOS  equal,  and  a  common  hypotenuse;  therefore 
the  sides  QR  and  QS  are  equal. 

Hence  the  two  bisectors  of  the  angles  between  the  two 
given  lines  constitute  the  locus  required. 

Ex.  Find  the  locus  of  a  point  from  which  the  perpendiculars  to 
two  given  parallel  lines  are  equal. 

265.  Problem  34.  To  find  the  locus  of  a  point  from 
which  the  perpendicular  to  a  given  line  shall  he  equal 
to  a  given  line-segment. 


Show  that  the  locus  consists  of  a  pair  of  lines  parallel  to  the  given 
line. 

Ex.  Find  the  locus  of  the  mid-points  of  all  the  lines  drawn  from  a 
given  point  to  a  given  line  not  passing  through  the  point. 

Intersection  of  Loci 

256.  Many  problems  relating  to  the  determination  of 
points  satisfying  given  conditions  can  be  solved  by  means  of 
the  intersection  of  loci. 

For  instance,  the  problem  to  determine  all  the  points  that 
satisfy  two  prescribed  conditions  A  and  B  may  be  solved  as 
follows : 

Construct  the  locus  of  the  points  satisfying  the  condition 
A ;  and  also  construct  the  locus  of  the  points  satisfying  the 
condition  B. 

The  points  of  intersection  of  the  two  loci  (and  these  points 
only)  satisfy  both  the  assigned  conditions. 


LOCUS  PROBLEMS  117 

257.   Problem  35.     To  find  a   point  such   that   its 
joins  to  three  given  points  not  collinear  are  equal. 

Let  A,  B,  c  be  the  three  given  points. 


"B 


It  is  required  to  find  a  point  0,  such  that  OA,  OB,  00  are 
all  equal. 

Draw  PP'  and  QQ'  the  perpendicular  bisectors  of  the  lines 
AB  and  BC. 

The  lines  PP'  and  QQ'  intersect,  for  if  they  were  parallel, 
then  the  lines  AB  and  BC,  being  respectively  perpendicular 
to  them,  would  be  in  one  straight  line.  Let  the  lines  PP' 
and  QQ'  intersect  in  0. 

To  prove  that  the  point  0,  and  no  other  point,  satisfies 
the  given  conditions. 

The  line  PP'  contains  all  those  points  and  only  those, 
whose  joins  to  A  and  B  are  equal  (253). 

The  line  QQ'  contains  all  those  points,  and  only  those, 
whose  joins  to  B  and  C  are  equal. 

Therefore  the  point  common  to  PP'  and  QQ',  and  no  other 
point,  has  its  joins  to  A,  B,  and  C  equal. 

Note.     This  construction  is  used  later  (III.  76)  in  finding  the  center 

of  a  given  circle. 

Ex.  1.  In  a  given  line  find  a  point  whose  joins  to  two  given  points 
are  equal. 

Ex.  2.  Find  a  point  from  which  the  perpendiculars  to  two  given 
intersecting  lines  are  respectively  equal  to  two  given  line-segments. 
Four  solutions  (255). 


118  PLANE  GEOMETRY — BOOK  I 

258.  Problem  36.  To  find  a  poiiit  from  which  the 
perpendiculars  to  three  given  lines  {forming  a  tri- 
angle) shall  he  equal. 

Let  the  three  lines  LL',  MM',  NN'  form  a  triangle  ABC. 


It  is  required  to  find  a  point  0,  such  that  the  perpendicu- 
lars from  0  to  these  lines  are  equal. 

[The  construction  and  proof  are  left  to  the  student.  Show  that 
there  are  four  solutions.] 

Ex.  1.  If  two  of  the  three  lines  are  parallel,  how  many  solutions 
are  there  ? 

Ex.  2.  On  a  given  line  find  those  points  from  which  the  perpen- 
diculars to  another  given  line  are  equal  to  an  assigned  line-segment. 

Ex.  3.  On  a  given  line  how  many  points  are  there  from  which  the 
perpendiculars  to  two  given  lines  are  equal  ? 

THEOREMS    ON    CONCURRENCE 

259.  Definition.  Three  or  more  lines  that  meet  in  a  com- 
mon point  (when  prolonged  if  necessary)  are  said  to  be  con- 
current. 

The  principle  of  the  intersection  of  loci  may  be  used  to 
prove  the  first  two  of  the  following  theorems  relating  to 
the  concurrence  of  certain  lines  in  a  triangle.  The  third  is 
then  derived  from  the  first  by  the  theory  of  parallels. 


LOCUS  PROBLEMS  119 

Concurrence  of  perpendicular  bisectors  of  sides, 

260.   Theorem  50.     In  any   triangle  the  three  per- 
pendicular bisectors  of  the  sides  are  concurrent. 

Outline.     Use  the  construction  of  267.     Then  show  that  0  lies  on 
the  perpendicular  bisector  of  the  side  AC. 


Concurrence  of  angle-bisectors, 

261.  Theorem  51.  In  any  triangle  the  three  bisec- 
tors of  the  interior  angles  are  concurrent. 

Outline.  Let  the  point  0  in  258  be  the  intersection  of  two  of  the 
angle-bisectors.     Show  that  0  lies  on  the  third  angle-bisector. 

262.  Cor.  Tlie  exterior  angle-bisectors  through  two  vertices 
and  the  interior  angle-bisector  through  the  third  vertex  are  con- 
current. 

263.  Definition.  The  lines  drawn  from  the  vertices  of  a 
triangle  perpendicular  to  the  opposite  sides,  respectively, 
are  called  the  principal  perpendiculars  of  the  triangle. 

Concurrence  of  principal  perpendiculars, 

264.  Theorem  52.  The  three  principal  perpendicu- 
lars of  a  triangle  are  concurrent. 

Let  ABC  be  any  triangle.  Let  AB,  BE,  and  CF  be  the 
principal  perpendiculars. 

C'  A        B' 


K' 

sX 

1 

>    /cr 

To  prove  that  AD,  BE,  and  CF  are  concurrent. 

MCM.  £LEM.  GEOM.  —  9 


120  PLANE  GEOMETRY — BOOK  I 

Through  the  vertices  A,  B,  C  draw  lines  parallel  respec- 
tively to  the  opposite  sides,  so  as  to  form  a  second  triangle 
A'B'C'. 

Outline.  Prove  by  153  that  A  is  the  mid-point  of  B'C,  B  the  mid- 
point of  A'C,  and  C  of  A'B' ;  hence  that  AD,  BE,  CF  are  the  per- 
pendicular bisectors  of  the  sides  of  the  new  triangle  A'B'C,  Then 
draw  desired  conclusion  and  quote  authority. 

On  Methods  of  Analysis 

265.  When  a  new  theorem  or  problem  is  presented  for  investiga- 
tion (as  in  the  miscellaneous  exercises  that  follow),  we  try  to  discover 
some  connection  or  relationship  between  the  new  proposition  and  the 
previous  ones  with  which  we  are  familiar.  This  relationship  is  to  be 
discovered  by  means  of  a  preliminary  analysis.  The  words  analysis 
and  synthesis  and  the  corresponding  adjectives  analytic  and  synthetic 
are  much  used  in  mathematics.  In  general,  analysis  means  the  sepa- 
ration of  a  whole  into  its  parts,  and  synthesis  means  bringing  the  parts 
together  to  make  a  whole.  In  geometry  the  words  are  used  in  a  more 
restricted  sense.  In  synthesis  we  begin  with  admitted  facts,  and,  by 
the  aid  of  principles  or  tlieorems  already  accepted  and  problems  al- 
ready solved,  we  prove  some  new  theorem  or  solve  some  new  problem. 
This  is  usually  the  most  convenient  way  of  presenting  the  result  when 
it  is  once  obtained  ;  but  the  actual  discovery  is  often  made  in  the 
reverse,  way  by  means  of  an  analysis,  in  which  we  begin  with  the  con- 
clusion and  then  examine  the  different  conditions  that  are  necessary 
or  sufficient  to  lead  to  the  result  in  question.  The  analysis  of  a  prob- 
lem was  described  in  131,  and  illustrated  in  various  subsequent  articles. 
The  analysis  of  a  theorem  is  somewhat  similar,  and  may  be  conducted 
in  two  ways,  which  may  be  called,  respectively,  the  analysis  of  ante- 
cedents and  the  analysis  of  consequents. 

266.  Analysis  of  antecedents.  In  this  method  we  examine 
the  antecedent  conditions  from  which  the  conclusion  in  question  would 
follow,  and  then  compare  these  conditions  with  the  given  hypothesis. 
For  example,  let  the  conclusion  be  called  '  statement  iS",'  then  the  analy- 
sis of  antecedents  may  be  put  in  the  following  form  :  — 

The  statement  S  is  true,  if  the  statement  B  is  true ; 
the  statement  R  is  true,  if  the  statement  Q  is  true ; 
the  statement  Q  is  true,  if  the  statement  P  is  true ; 


ANALYSIS  121 

and  so  on.  If  by  this  method  we  get  back  to  some  antecedent  state- 
ment A  which  we  know  to  be  true  by  some  principle  already  accepted, 
or  which  would  follow  from  the  given  hypothesis,  then  we  are  war- 
ranted in  asserting  the  truth  of  statement  iS.  The  successive  steps 
from  A  to  S  can  then  be  presented  in  the  reverse  of  the  order  just 
given,  and  the  proof  can  be  arranged  in  the  usual  synthetic  form 
beginning  with  the  hypothesis  and  ending  with  the  conclusion  to  be 
demonstrated. 

If,  however,  we  come  only  to  a  statement  that  we  know  to  be  false 
(or  do  not  know  to  be  true) ,  then  the  statement  S  may  or  may  not  be 
true,  and  nothing  is  proved.  A  new  set  of  antecedent  conditions  may 
then  be  examined.  This  method  often  proves  the  truth  of  a  theorem  ; 
it  cannot  by  itself  prove  any  statement  false. 

267.  Analysis  of  consequents,    in  this  method  we  examine 

the  consequences  that  would  follow  if  the  theorem  were  supposed  to  be 
true,  and  then  compare  these  consequences  vsrith  the  hypothesis  and 
other  accepted  facts.  The  analysis  of  consequents  may  be  put  in  the 
following  form : 

If  the  statement  S  is  true,  then  the  statement  T  is  true ; 
if  the  statement  T  is  true,  then  the  statement  U  is  true  ; 

and  so  on.  If  by  this  method  we  arrive  at  some  statement  that  we 
know  to  be  false  (or  inconsistent  with  the  hypothesis),  then  we  con- 
clude that  the  statement  S  is  false,  since  it  can  be  reduced  to  an 
absurdity. 

If,  however,  we  come  only  to  a  statement  Z  that  we  know  to  be  true 
(or  do  not  know  to  be  false),  then  the  statement  S  may  or  may  not  be 
true,  and  nothing  is  proved.  This  method  of  analysis  often  proves  the 
falsity  of  a  statement ;  it  cannot  by  itself  prove  any  statement  true, 
since  the  steps  taken  from  S  to  Z  are  not  always  reversible  ;  it  some- 
times, however,  points  the  way  to  a  synthetic  proof  by  reversal  of  the 
steps. 

268.  Analysis  of  the  opposite.  Either  of  the  two  methods 
of  analysis  may  be  applied  to  the  opposite  of  the  statement  S.  The 
analysis  of  antecedents  gives  a  decisive  result  if  we  arrive  at  an  ante- 
cedent known  to  be  true,  for  then  the  opposite  of  S  is  true,  and  ^  is 
false.  The  analysis  of  consequents  gives  a  decisive  result  if  we  arrive 
at  a  consequent  known  to  be  false,  for  then  the  opposite  of  8  is  false, 
and  S  is  true  ;  this  case  is  the  familiar  reductio  ad  absurdum  of  which 
several  illustrations  have  been  given  (see  102). 


122  PLANE  GEOMETRY  —  BOOK  I 

EXERCISES  ON    BOOK  I 

1.  In  an  isosceles  triangle,  if  a  perpendicular  is  drawn  from  an 
extremity  of  the  base  to  the  opposite  side,  then  the  angle  between  this 
perpendicular  and  the  base  is  equal  to  half  the  vertical  angle. 

2.  In  a  right  triangle,  prove  that  a  line  can  be  drawn  dividing  the 
right  angle  into  two  parts  equal  respectively  to  the  other  angles,  and  so 
as  to  divide  the  right  triangle  into  two  isosceles  triangles. 

3.  In  a  right  triangle  the  median  drawn  to  the  mid-point  of  the 
hypotenuse  equals  half  the  latter. 

4.  Through  two  given  points  draw  two  lines  forming  with  a  given 
indefinite  line  an  equilateral  triangle.    How  many  solutions  are  there  ? 

5.  The  bisectors  of  the  base  angles  of  an  isosceles  triangle  contain 
an  angle  equal  to  an  exterior  angle  of  the  triangle. 

6.  The  lines  joining  the  adjacent  extremities  of  unequal  and  paral- 
lel lines  will  meet  if  prolonged  through  the  extremities  of  the  shorter 
parallel  (124). 

7.  Construct  a  right  triangle,  being  given  the  hypotenuse  and 
the  sum  (or  difference)  of  the  two  sides  (137,  ex.  1).  Construct  a 
right  triangle,  being  given  one  side  and  the  sum  (or  difference)  of  the 
other  side  and  the  hypotenuse  (137,  ex.  3). 

8.  Lines  drawn  from  two  opposite  vertices  of  a  parallelogram  to 
the  mid-points  of  a  pair  of  opposite  sides  trisect  a  diagonal  (167). 

9.  Lines  drawn  from  any  vertex  of  a  parallelogram  to  the  mid- 
points of  the  two  non-adjacent  sides  trisect  a  diagonal  (178). 

10.  If  alternate  sides  of  a  pentagon  are  prolonged  to  meet,  then 
the  sum  of  the  five  angles  so  formed  is  equal  to  two  right  angles  (190). 

11.  If  alternate  sides  of  a  hexagon  are  prolonged  to  meet,  then  the 
sum  of  the  six  angles  so  formed  is  equal  to  four  right  angles.  Consider 
also  the  general  case  of  an  n-gon. 

12.  Through  a  given  point  draw  a  line  so  that  the  part  intercepted 
between  two  given  parallel  lines  may  be  equal  to  a  given  line. 

13.  Through  a  given  point  within  a  given  fixed  angle  draw  a  line 
so  that  the  segment  between  the  sides  may  be  bisected  at  the  point 
(186,  ex.  6). 

14.  Construct  a  triangle,  being  given  two  sides  and  the  median 
drawn  to  the  mid-point  of  the  third  side. 

15.  Construct  a  triangle  being  given  one  side  and  the  medians  to 
the  mid-points  of  the  other  two  sides. 


EXERCISES  123 

16.  Any  line  through  the  intersection  of  the  diagonals  of  a  paral- 
lelogram and  terminated  by  opposite  sides  is  bisected  at  that  point. 

17.  In  a  given  triangle  inscribe  a  parallelogram  having  one  side 
resting  on  the  base,  and  having  the  intersection  of  its  diagonals  at  a 
given  point. 

18.  The  bisectors  of  the  angles  of  a  parallelogram  form  a  rectangle 

(123). 

19.  The  bisectors  of  the  angles  of  a  rectangle  form  a  square. 

20.  In  an  isosceles  triangle  the  bisector  of  a  base  angle,  and  the 
bisector  of  the  external  angle  supplemental  to  the  other  base  angle, 
form  an  angle  equal  to  half  the  vertical  angle. 

21.  Construct  a  triangle,  being  given  the  angles,  and  one  of  the 
principal  perpendiculars. 

22.  Construct  a  triangle,  being  given  the  mid-points  of  the  three 
sides  (see  figure  in  264). 

23.  Draw  a  parallel  to  the  base  of  a  triangle,  so  that  the  intercept 
may  be  equal  to  one  of  the  segments  adjacent  to  the  base  (compare 
186,  ex.  5). 

24.  Draw  a  parallel  to  the  base  of  a  triangle,  so  that  the  intercept 
may  be  equal  to  the  sum  of  the  segments  adjacent  to  the  base. 

25.  Draw  a  parallel  to  a  base  of  the  triangle,  so  that  the  intercept 
may  be  equal  to  the  difference  of  the  segments  adjacent  to  the  base. 

26.  Given  the  sum  (or  difference)  of  the  side  and  principal  per- 
pendicular of  an  equilateral  triangle,  construct  it  (137,  ex.  6).  ^ 

27.  Given  the  sum  (or  difference)  of  the  side  and  diagonal  of  a 
square,  construct  it. 

28.  If  the  opposite  sides  of  a  hexagon  are  parallel,  then  its  diagonals 
are  concurrent. 

29.  Given  two  indefinite  lines  and  a  point :  to  find  a  point  in  one  of 
the  lines  so  that  the  line  joining  it  to  the  given  point  may  be  bisected 
by  the  other  line.    How  many  solutions  are  there  ? 

30.  If  through  any  vertex  of  a  parallelogram  a  line  is  drawn,  and 
if  perpendiculars  to  this  line  are  drawn  from  the  other  vertices,  then 
the  perpendicular  from  the  vertex  opposite  the  first  is  equal  to  the 
sum  or  difference  of  the  other  two,  according  as  the  line  passes  without 
or  within  the  parallelogram. 

31.  In  any  quadrangle  the  two  lines  joining  the  mid-points  of  oppo- 
site sides,  and  the  line  joining  the  mid-points  of  the  diagonals,  all  meet 
in  a  point  and  bisect  each  other  (175,  ex.  3). 


124  PLANE  GEOMETRY — BOOK  I 

32.  If  three  parallel  lines  make  equal  intercepts  on  a  transversal, 
and  if  a  second  transversal  cross  the  first  between  two  of  the  parallels, 
then  the  intercept  on  the  middle  parallel  equals  half  the  difference  of 
the  intercepts  on  the  other  two  (compare  172). 

33.  If  through  the  extremities  of  the  base  of  a  triangle  whose  sides 
are  unequal,  lines  are  drawn  to  any  point  in  the  bisector  of  the  vertical 
angle,  their  differences  are  less  than  the  difference  of  the  sides. 

[Let  side  AB  be  greater  than  AC.  On  AB  take  AC  equal  to  AC. 
Join  C  to  the  point  in  the  bisector.] 

34.  If  the  lines  in  ex.  33  are  drawn  to  any  point  in  the  bisector 
of  the  external  vertical  angle,  then  their  sum  is  greater  than  the  sum 
of  the  sides. 

35.  If  one  of  the  acute  angles  of  a  right  triangle  is  double  the  other, 
the  hypotenuse  is  double  the  shortest  side. 

36.  If  a  quadrangle  is  inscribed  in  a  parallelogram  and  has  its  oppo- 
site vertices  symmetric  as  to  the  center  of  symmetry  of  the  parallelo- 
gram (244),  then  the  quadrangle  is  a  parallelogram. 

37.  If  a  parallelogram  is  inscribed  in  a  rectangle,  having  its  sides 
parallel  to  the  diagonals  of  the  rectangle,  then  two  adjacent  sides  of 
the  parallelogram  make  equal  angles  with  a  side  of  the  rectangle. 

38.  A  billiard  ball  is  placed  at  any  point  of  a  rectangular  table. 
In  what  direction  must  it  be  struck  so  that  it  shall  return  to  the  first 
point  after  being  reflected  successively  at  the  four  sides,  the  lines  of 
motion,  before  and  after  impact,  making  equal  angles  with  the  suc- 
cessive sides  of  the  table  ? 

39.  If  two  sides  of  a  triangle  are  unequal  then  the  bisector  of  the 
angle  between  them  divides  the  opposite  side  into  unequal  segments, 
the  greater  segment  being  adjacent  to  the  greater  side. 

Outline.  In  triangle  ABC,  let  AD  bisect  the  angled.  Given  AB 
'greater  than  ^C ;  to  prove  BD  greater  than  BC.  On  AB  lay  off  AC 
equal  to  AC,  and  join  CD.  By  equality  and  inequality  of  angles 
prove  angle  B  less  than  BCD.     Draw  conclusion. 

40.  To  inscribe  a  square  in  a  right  triangle. 

41.  To  inscribe  a  square  in  a  rhombus. 

42.  If  two  isosceles  triangles  have  equal  bases,  and  if  one  of  the 
equal  sides  of  the  first  triangle  is  greater  than  one  of  the  equal  sides  of 
the  second,  then  the  vertical  angle  of  the  first  triangle  is  less  than  the 
vertical  angle  of  the  second. 


BOOK  II,  — EQUIVALENCE  OF  POLYGONS 

GENERAL  PRINCIPLES 

1.  Definitions.  Two  polygons  are  said  to  be  joined  when 
they  are  brought  together,  without  overlapping,  so  that  a 
side  of  one  coincides  in  whole  or  in  part  with  a  side  of  the 
other. 


When  the  common  portion  of  the  boundaries  of  two  joined 
polygons  is  erased  (or  ignored),  the  third  polygon  so  formed 
is  called  the  sum  of  the  two  original  polygons,  which  are 
then  said  to  be  added  together. 

A  polygon  is  said  to  be  dissected  when  its  surface  is 
divided  np  into  any  number  of  smaller  polygons  by  drawing 
straight  lines. 

Two  polygons  are  called  equivalent  if  their  surfaces  can 
be  dissected  so  that  each  part  of  one  is  separately  super- 
posable  on  some  part  of  the  other  by  suitable  rearrangement 
of  parts  if  necessary. 


F 


126  PLANE  GEOMETRY — BOOK  II 

Thus  the  triangle  ABC  and  the  rectangle  DEFG  are  equiva- 
lent if  the  parts  marked  with  corresponding  numerals  are 
superposable. 

To  make  this  definition  of  the  equivalence  of  polygons  consistent 
with  itself,  it  is  necessary  to  prove  the  following  two  lemmas  relating 
to  the  permanence  of  such  equivalence. 

2.  Lemma  1.  If  one  polygon  incloses  another  within  its  boundary 
so  that  the  latter  is  part  of  the  former,  it  is  not  possible  to  dissect  the 
inner  polygon  and  then  rearrange  and  join  its  parts  in  such  a  way  as 
to  cover  the  whole  of  the  outer  polygon. 

For  suppose  that  this  operation  is  possible  ;  and  let  the  outer  poly- 
gon be  supposed  actually  covered  by  the  rearranged  parts  of  the  inner 
one.  Remove  the  excess  of  the  outer  polygon  over  the  original  inner 
one.  Dissect  the  remaining  inner  polygon  as  before  and  then  rearrange 
the  parts  so  as  to  cover  the  outer  polygon.  Remove  the  excess  again  ; 
and  repeat  the  process  as  often  as  desired.  The  excess  can  be  accumu- 
lated until  it  is  more  than  sufficient  to  cover  any  polygon  however 
large.  But  this  excess  is  only  a  part  of  the  original  surface  of  the 
inner  polygon.  Therefore  the  surface  of  this  finite  polygon  can  be  so 
rearranged  as  to  cover  an  indefinitely  extended  surface,  which  is  absurd. 
Hence  the  lemma  is  established. 

3.  Lemma  2.  If  two  polygons  are  equivalent  for  one  mode  of  dissec- 
tion and  superposition,  they  will  be  equivalent  for  all  possible  modes 
of  dissection  and  superposition. 


Let  the  polygons  A  and  B  be  such  that  there  is  one  way  of  dissecting 
them  so  that  every  part  of  A  can  be  fitted  on  an  equal  and  corre- 
sponding part  of  jB,  the  latter  polygon  being  then  just  covered  by  the 
parts  of  the  former. 


GENERAL   PRINCIPLES  127 

Next  let  the  polygon  A  be  dissected  in  any  second  way,  and  let  the 
parts  be  placed  in  any  order  upon  B,  these  smaller  polygons  being 
joined  so  as  not  to  overlap.  Should  portions  of  any  of  them  extend 
over  the  boundary  of  B,  let  the  surplus  be  cut  off,  and  then  used  to 
cover  any  uncovered  portion  of  the  surface  of  B.  Continue  this  process 
until  either 

(a)  the  parts  of  A  are  exhausted,  leaving  a  portion  of  B  uncovered,  or 
(/3)  the  surface  of  B  is  covered,  leaving  a  portion  of  A  extending  over 
the  boundary  of  B,  or 

(7)  the  surface  of  B  is  just  covered  by  the  parts  of  A  without  excess 
or  defect. 

In  case  (a)  let  the  second  mode  of  dissection  and  superposition  cover 
B',  leaving  a  portion  C  uncovered.  Dissect  this  covering  of  B'  by  the 
second  mode  and  fit  the  parts  back  so  as  to  form  the  original  polygon 
A.  Then  dissect  A  by  the  original  mode,  and  rearrange  the  parts  so 
as  to  cover  the  polygon  B  (in  accordance  with  the  hypothesis).  Thus 
the  surface  of  the  inner  polygon  B'  has  been  rearranged  to  cover  the 
surface  of  the  outer  polygon  B,  contrary  to  the  preceding  lemma ;  hence 
case  (a)  cannot  occur. 

In  case  (^)  let  the  second  mode  of  superposition  cover  a  polygon 
made  up  of  B  and  the  surplus  D.  Dissect  this  covering  of  B  by  the 
first  mode,  and  use  the  parts  to  form  the  polygon  A  (in  accordance 
with  the  hypothesis);  then  dissect  A  by  the  second  mode,  and  cover 
the  polygon  made  up  of  B  and  D.  Thus  the  surface  of  the  inner 
polygon  B  has  been  rearranged  to  cover  the  outer  polygon,  contrary  to 
the  preceding  lemma  ;  hence  case  (/3)  cannot  occur. 

Therefore  case  (7)  is  the  only  one  that  can  occur ;  that  is  to  say, 
the  polygon  B  is  just  covered  by  the  second  mode  of  dissecting  A  and 
of  superposing  the  parts  on  B. 

4.  Definitions  continued.  One  polygon  is  said  to  be  greater 
than  a  second  polygon  if  a  portion  of  the  first  can  be  dis- 
sected and  rearranged  so  as  to  cover  the  second.  In  the 
same  case  the  second  is  said  to  be  less  than  the  first. 

5.  Any  polygon  whieh,  when  added  to  the  less  of  two 
given  polygons,  forms  a  polygon  equivalent  to  the  greater, 
is  called  the  difference  of  the  two  given  polygons. 

6.  If  two  polygons  are  equivalent,  any  polygon  equivalent 
to  their  sum  is  said  to  be  the  double  of  either  polygon,  and 
each  of  the  former  is  said  to  be  equivalent  to  Judf  the  latter. 


128  PLANE  GEOMETRY  —  BOOK  II 

Axioms  op  Equivalence  and  Non-equivalence* 

From  the  foregoing  definitions,  the  following  statements 
are  direct  inferences  by  means  of  the  principle  of  superposi- 
tion: 

7.  Polygons  which  are  equivalent  to  the  same  poly- 
gon are  equivalent  to  each  other. 

%.  If  a  nurnber  of  polygons  are  added  together  in  a 
certain  way  and  order,  the  sum  is  equivalent  to  the  sum 
that  would  have  been  obtained  if  the  polygons  had  been 
added  together  in  a  different  way  or  in  a  different 
order. 

For  the  two  resulting  polygons  can  be  dissected  into  superposable 
parts. 

9.  If  equivalent  polygons  are  added  to  equiva- 
lent polygons,  the  sums  are  equivalent  polygons.  In 
particular  the  doubles  of  equivalent  polygons  are 
equivalent. 

10  (a).  //  two  unequivalent  polygons  are  added  re- 
spectively to  unequivalent  polygons,  tlie  sum  of  the  two 
greater  polygons  is  greater  than  the  sum  of  tlie  two 
less  ones. 

10  (^).  If  one  polygon  is  greater  than  a  second,  the 
double  of  the  first  is  greater  than  the  double  of  the 
second. 

1 1 .  The  halves  of  equivalent  polygons  are  equivalent. 

Apply  indirect  proof  and  use  10  (6). 

12.  If  one  polygon  is  greater  than  a  second,  then  the 
half  of  the  first  polygon  is  greater  than  the  half  of  the 
second. 

*  The  student  need  not  dwell  on  Arts.  7-17  at  first  reading,  but 
should  refer  back  to  them  when  necessary. 


GENERAL  PRINCIPLES 


129 


General  Theorems  relating  to  Equivalence 

13.  Theorem  1.  The  double  of  the  sum  of  two  poly- 
gons is  equivalent  to  the  sum  of  the  doubles  of  the  two 
polygons. 

For  the  double  of  the  sum  of  two  polygons  A  and  -B  is  a 
polygon  made  up  of  the  four  parts  A,  B,  A,  B;  and  the  sum 
of  the  doubles  of  A  and  5  is  a  polygon  made  up  of  the  four 
parts  A,  A,B,  B;  differing  only  in  the  order  of  arrangement; 
hence  the  two  sums  are  equivalent  (8). 

14.  Theorem  2.  The  half  of  the  sum^  of  two  poly gons 
is  equivalent  to  the  sum  of  the  halves  of  the  polygons. 

For  the  sum  of  the  halves  when  doubled  becomes  equiva- 
lent to  the  sum  of  the  two  whole  polygons  (13),  and  is  there- 
fore, by  definition,  equivalent  to  half  this  sum. 

15.  Theorem  3.  //  equivalent  polygons  are  taken 
away  from,  equivalent  polygons,  the  remaining  figures 
are  equivalent. 


Let  the  polygons  A  and  B  be  equivalent;  and  let  the 
polygons  C  and  D  be  equivalent ;  then  the  remaining  figures 
M  and  N  are  equivalent. 

For,  by  Lemma  2,  A  can  be  fitted  on  J5  by  any  mode  of 
dissection.  Choose  a  mode  in  which  the  parts  of  C  are  made 
to  cover  its  equivalent  D.  Then  the  parts  of  3/  will  cover 
the  remaining  figure  N. 

15  («).  Ex.  Prove  the  axioms  of  non-equivalence  relating  to  sub- 
traction.    (See  ax.  10,  11,  p.  17.) 


130  PLANE  GEOMETRY — BOOK  II 

16.  Theorem  4.  The  double  of  the  difference  of  two 
polygons  is  equivalent  to  the  difference  of  their  doubles. 

For  let  the  difference  of  the  polygons  A  and  B  be  the  poly- 
gon C;   then  the  sum  of   B  and  C  is  equivalent  to  A  (5). 

Therefore  the  sum  of  the  doubles  of  B  and  C  is  equivalent 
to  the  double  of  A  (13);  hence  the  difference  between  the 
double  of  A  and  the  double  of  B  is  equivalent  to  the  double 
of  C  (5),  and  is  therefore  equivalent  to  double  the  difference 
between  A  and  B. 

17.  Theorem  5.  The  half  of  the  difference  of  two 
polygons  is  equivalent  to  the  difference  of  their  halves. 

Use  a  similar  proof,  substituting  the  word  half  for  the  word  double. 

COMPARISON   OF   PARALLELOGRAMS 

18.  Definitions.  In  a  given  triangle  the  line  drawn  from 
any  vertex  perpendicular  to  the  opposite  side  is  called  the 
altitude  of  the  triangle  with  reference  to  that  sicle  taken  as 
base.  Any  side  may  be  so  regarded  as  base  and  the  corre- 
sponding perpendicular  as  the  altitude;  hence  a  triangle 
has  three  altitudes. 

Similarly  any  side  of  a  parallelogram  may  be  regarded  as 
its  base,  and  the  line  drawn  perpendicular  to  it  from  any 
point  of  the  opposite  side  may  be  taken  as  corresponding 
altitude.  All  such  altitudes  drawn  to  the  same  side  are 
equal,  and  are  also  equal  to  the  altitudes  drawn  to  the 
opposite  side.     Thus  a  parallelogram  has  only  two  altitudes. 


19.    In  the  case  of  a  rectangle,  when  any  particular  side 
is  taken  as  base,  either  of  the  adjacent  sides  is  the  altitude. 


COMPARISON  OF  PARALLELOGRAMS 


131 


20.  A  rectangle  is  completely  determined  by  two  adjacent 
sides;  that  is  to  say,  all  the  rectangles  whose  adjacent  sides 
are  equal  to  two  given  lines  are  superposable  (I.  166) ;  and, 
for  this  reason,  each  of  these  rectangles  is  called  the  rec- 
tangle of  the  two  given  lines. 


Rectangles  of  equal  altitudes, 

21.  Theorem  6.  If  two  rectangles  have  equal  alti- 
tudes and  unequal  bases,  that  which  hojS  the  greater 
base  is  the  greater  rectangle. 

Let  the  rectangles  ABCD,  EFGH  have  their  altitudes  AB 
and  EF  equal.     Let  the  base  AD  he  greater  than  the  base  EH. 


L   G 


K  D 


H 


To  prove  that  ABCD  is  greater  than  EFGH. 

Lay  off  ^^  equal  to  EH,  and  complete  the  rectangle  ABLE. 

This  rectangle  is  equal  to  EFGH  (I.  166).  Hence  a  por- 
tion of  ABCD  will  cover  EFGH.  Therefore  ABCD  \s  greater 
than  EFGH  (4). 

22.  Cor.  I.  If  two  rectangles  have  equal  altitudes,  then 
according  as  the  base  of  the  first  is  greater  than,  equal  to,  or 
less  than  the  base  of  the  second,  so  is  the  first  rectangle  greater 
than,  equal  to,  or  less  than  the  second.     (21 ;  and  I.  166.) 

23.  Cor.  2.  If  two  rectangles  have  equal  altitudes,  then 
according  as  the  first  rectangle  is  greater  thxn,  equal  to,  or  less 


132 


PLANE  GEOMETRY— BOOK  II 


than  the  second,  so  is  the  base  of  the  first  greater  than,  equal  to, 
or  less  than  the  base  of  the  second.    (Rule  of  Conversion,  1. 104.) 

Ex.  Show  that  22  and  23  are  still  true  if  the  words  base  and  altitude 
are  interchanged  throughout. 

24.  Cor.  3.  According  as  the  side  of  one  square  is  greater 
than,  equal  to,  or  less  than  the  side  of  another  square,  so  is  the 
first  square  greater  than,  equal  to,  or  less  than  the  second;  and 
conversely. 

The  student  may  give  an  independent  proof  by  superposition  ;  and 
then  state  the  converse. 

JParallelogratns  and  rectangles, 

25.  Theorem  7.  A  parallelogranv  is  equivalent  to 
the  rectangle  of  its  base  and  altitude. 

Let  ABCD  be  the  given  parallelogram,  having  ^5  for  base 
and  AF  OT  BE  for  altitude. 


To  prove  that  ABCD  is  equivalent  to  ABEF,  the  rectangle 
of  its  base  and  altitude. 

In  the  triangles  AFD  and  BEG:  the  side  AF  equals  BE 
(I.  153) ;  the  side  AD  equals  BC;  and  the  angle  FAD  equals 
EBC,  having  parallel  sides  (I.  126). 

Therefore  the  triangles  AFD  and  BEC  are  equal. 

Take  these  equivalents  in  turn  away  from  the  quadrilateral 
ABCF;  then  the  remainders  ABCD  and  ABEF  are  equiva- 
lent (15). 

Ex.  Prove  the  theorem  for  the  case  in  which  D  and  E  coincide. 


COMPARISON  OF  PARALLELOGRAMS  133 

26.    Cor.  I.     Two  parallelograms  having  equal  bases  and 
equal  altitudes  are  equivalent. 

Ex.    Show  how  to  dissect  any  two  parallelograms  having  the  same 
base  and  equal  altitudes,  so  that  the  parts  may  be  superposable. 


27.  Cor.  2.  If  two  parallelograms  have  equal  altitudes^ 
then  according  as  the  base  of  the  first  is  greater  than,  equal  to, 
or  less  than  the  base  of  the  second,  so  is  the  first  parallelogram 
greater  than,  equivalent  to,  or  less  than  the  second.  (Use  22 
and  25.) 

28.  Cor.  3-  If  two  parallelograms  have  equal  altitudes,  then 
according  as  the  first  parallelogram  is  greater  than,  equivalent 
to,  or  less  than  the  second,  so  is  the  base  of  the  first  greater 
than,  equal  to,  or  less  than  the  base  of  the  second.  (Rule  of 
Conversion.) 

Ex.  Show  that  27  and  28  are  still  true  if  the  words  base  and  altitude 
are  interchanged  throughout. 

Triangles  and  rectangles, 

29.  Theorem  8.  A  triangle  is  equivalent  to  half  the 
rectangle  of  its  base  and  altitude. 


AD  B 


Let  ABC  be  the  triangle,  having  AB  for  base  and  CD  for 
altitude. 


134 


PLANE  GEOMETRY — BOOK  II 


To  prove  that  ABCi^  equivalent  to  the  rectangle  oi  AB,  CD. 

Complete  the  parallelogram  ABEC. 

The  triangle  ABC  is  equivalent  to  half  the  parallelogram 
ABEC  (1. 153)  ;  and  therefore  equivalent  to  half  the  rectangle 
oi  AB  and  CD  (25). 

p]x.  Prove  this  theorem  directly  by  applying  14,  17  to  the  adjoin- 
iug  figures. 


30.  Cor.  I.  A  trapezoid  is  equivalent  to  the  rectangle  con- 
tained by  its  altitude  and  half  tJie  sum  of  its  parallel  sides. 

31.  Cor.  2.  If  two  triangles  have  equal  altitudeSf  then 
accordiiig  as  the  base  of  the  first  is  greater  thaiiy  equal  to,  or 
less  than  the  base  of  the  second,  so  is  the  first  triangle  greater 
than,  equivalent  to,  or  less  than  the  second.     (Use  23,  29.) 

32.  Cor.  3.  If  two  triangles  have  equal  altitudes,  then 
according  as  the  first  tnangle  is  greater  than,  equivalent  to,  or 
less  than  the  second,  so  is  the  base  of  the  first  greater  than, 
equivalent  to,  or  less  than  the  base  of  the  second. 

Ex.  If  there  are  two  equilateral  triangles,  then  according  as  a  side 
of  the  first  is  greater  than,  equal  to,  or  less  than  a  side  of  the  second, 
so  is  the  first  triangle  greater  than,  equal  to,  or  less  than  the  second. 

33.  Cor.  4.  Two  triangles  having  the  same  base,  and  having 
their  opposite  vertices  on  the  same  line  parallel  to  the  base,  are 
equivalent.  Conversely,  two  equivalent  triangles  on  the  same 
base  and  at  the  same  side  of  it  are  between  the  same  parallels. 


34.  Cor.  5.  If  a  parallelogram  and  a  triangle  are  upon  the 
same  base  and  betiveen  the  same  parallels,  the  parallelogram  is 
double  the  triangle. 


COMPARISON  OF  PARALLELOGRAMS  135 

Parallelograms  about  a  diagonal, 

35.  Definition.  If  through  any  point  on  the  diagonal  of  a 
parallelogram  two  lines  be  drawn  parallel  to  the  sides,  so  as 
to  divide  the  parallelogram  into  four  smaller  parallelograms, 
the  two  whose  diagonals  are  portions  of  the  diagonal  first 
mentioned  are  called  the  parallelograms  about  the  diago- 
nal ;  and  the  two  which  lie  one  on  each  side  of  the  diagonal 
are  called  the  complements  of  the  parallelogram,s  about 
the  diagonal. 

36.  Theorem  9.  The  complements  of  the  parallelo- 
grams  about  the  diagonal  of  a  paraZlelogramj  are 
equivalent. 

Let  ABCD  be  the  parallelogram,  BD  its  diagonal,  K  any 
point  on  it,  FH  and  EG  lines  through  K  parallel  to  the  sides, 
forming  KGBF  and  DHKE  parallelograms  about  the  diagonal, 
and  KFAE  and  CGKH  the  complements  of  these  parallelo- 
grams. 


A  F        B 


To  prove  that  these  complements  are  equivalent. 

Since  the  diagonal  of  a  parallelogram  bisects  it  (I.  153), 
the  triangle  DBA  is  equivalent  to  CBD ;  similarly  KBF  is 
equivalent  to  KGB  ;  and  RED  to  DHK. 

Take  KBF  and  KED  away  from  DBA,  and  take  KGB  and 
DHK  away  from  CBD  ;  then  the  remainders  KFAE  and  KHCQ 
are  equivalent  (15). 

Note.  This  theorem  is  useful  in  the  construction  of  equivalent 
parallelograms  (72). 

MCM.  ELEM.  GEOM. 10 


136  PLANE  GEOMETRY — BOOK  II 

37.  Theorem  10.  Parallelograms  about  the  diagonal 
of  a  rhombus  are  rhombuses,  and  their  complements 
are  equal  parallelograms. 


Use  161,  117,  166  of  Book  I.    . 

38.   Cor.     Parallelograms  about  the  diagonal  of  a  square 
are  squareSj  and  their  complements  are  equal  rectangles. 


EXERCISES 

1.  If  one  diagonal  of  a  quadrangle  bisects  the  other,  it  also  bisects 
the  quadrangle. 

2.  If  a  parallelogram  and  a  triangle  are  such  that  the  base  and 
altitude  of  the  parallelogram  are  respectively  equal  to  half  the  base 
and  altitude  of  the  triangle,  then  the  parallelogram  is  equivalent  to 
half  the  triangle. 

3.  Lines  joining  the  mid-points  of  adjacent  sides  of  a  quadrangle 
form  a  parallelogram  equivalent  to  half  the  quadrangle. 

4.  If  two  triangles  stand  on  the  same  base  and  at  the  same  side  of 
it,  and  if  the  middle  points  of  the  sides  are  joined,  then  the  joining  lines 
form  a  parallelogram  equivalent  to  half  the  difference  of  the  triangles. 

5.  To  construct  an  isosceles  triangle  equivalent  to  a  given  triangle 
and  standing  on  the  same  base. 

6.  To  construct  a  rhombus  equivalent  to  a  given  parallelogram  and 
having  the  same  diagonal. 

7.  A  triangle  whose  base  is  one  of  the  non-parallel  sides  of  a  trape- 
zoid and  whose  vertex  is  at  the  mid-point  of  the  opposite  side  is  equiva- 
lent to  half  the  trapezoid. 

[Through  the  mid-point  in  question  draw  a  parallel  to  the  oppo- 
site side  and  complete  the  parallelogram.] 


EQUIVALENCES    INVOLVING    RECTANGLES     137 

EQUIVALENCES   INVOLVING  RECTANGLES 
Rectangles  of  wholes  and  parts, 

39.  Theorem  11.  If  there  are  two  lines,  one  of  which 
is  divided  into  any  number  of  parts  at  ^iven  points, 
the  rectangle  of  the  two  given  lines  is  equivalent  to 
the  sum  of  the  rectangles  of  the  undivided  line  and  the 
several  parts  of  the  divided  line. 

Let  ABy  CF  be  the  two  lines,  and  let  CF  be  divided  at 
the  points  D  and  E  into  the  parts  C7i),  DEy  EF. 


To  prove  that  the  rectangle  oi  AB  and  CF  is  equal  to  the 
sum  of  the  rectangles  of  AB,  CD  ;  AB,  DE ;  AB,  EF. 

Draw  the  line  CG  perpendicular  to  CF  and  equal  to  AB. 
Complete  the  rectangle  CFLG,  and  draw  DH,  EK  perpendicu- 
lar to  CF. 

The  lines  DH,  EK  are  equal  to  CG  (I.  153)  and  therefore 
equal  to  AB. 

The  rectangle  CL  is  equivalent  to  the  sum  of  the  rec- 
tangles CH,  DK,  EL. 

Now  CH  is  the  rectangle  of  CG  and  CD,  that  is,  of  AB  and 
CD ;  also  DK  is  the  rectangle  of  AB  and  DE]  and  EL  is  the 
rectangle  of  AB  and  EF. 

Therefore  the  theorem  is  established. 


Note.     Two  of  the  following  corollaries  are  special 
theorem,  and  the  third  is  an  extension  of  it. 


cases  of  this 


138  PLANE  GEOMETRY — BOOK  II 

Rectangle  of  whole  line  and  one  parU 

40  (a).  Cor.  i.  If  a  line  is  divided  into  any  two  partSy  the 
rectangle  of  the  whole  line  and  one  part  is  equivalent  to  the 
square  on  that  part  together  with  the  rectangle  of  the  two  parts. 


Square  on  whole  line, 

40  (6).  Cor.  2.  If  a  line  is  divided  into  any  two  parts,  the 
square  on  the  ivhole  line  is  equivalent  to  the  sum  of  the  rec- 
tangles of  the  ivhole  line  and  each  of  the  parts. 


Distributive  property  of  rectangles, 

41.  Cor.  3.  If  each  of  two  lines  is  divided  into  any  number 
of  parts,  then  the  rectangle  contained  by  the  whole  lines  is  equiv- 
alent to  the  S7tm  of  all  the  rectangles  contained  by  each  part  of 
one  and  each  part  of  the  other. 

[Prove  by  repeated  applications  of  39 ;  or  else  by  an  independent 
figure.] 

Note.  This  important  principle  will  be  referred  to  as  "  the  dis- 
tributive property  of  rectangles"  ;  it  lies  at  the  foundation  of  many 
of  the  subsequent  theorems. 

Ex.  1.  Show  that  39,  40  are  special  cases  of  the  "distributive 
property." 

Ex.  2.  If  a  line  is  divided  into  three  parts,  then  the  square  on  the 
whole  line  is  equivalent  to  the  sum  of  the  rectangles  of  the  whole  line 
and  each  of  its  parts. 


EQUIVALENCES  INVOLVING  RECTANGLES     139 

Squares  on  whole  and  parts, 

42.  Theorem  12.  //  a  line  is  divided  into  any  two 
parts,  the  square  on  the  whole  line  is  equivalent  to 
the  sum  of  the  squares  of  the  parts  and  double  the 
rectangle  contained  hy  the  parts. 

Let  AB  be  the  given  line  divided  at  E. 
V  G        n 


F 


A  E  B 

To  prove  that  the  square  on  AB  is  equivalent  to  the  sum 
of  the  squares  on  AE  and  EBj  and  twice  the  rectangle  of  AE 
and  EB. 

[Use36,  37,  38.] 

Symbolic  Proof.  Another  simple  proof  may  be  given  by  using 
the  distributive  property  of  rectangles.  For  brevity  denote  the  rec- 
tangle of  two  lines  AB  and  CD  by  [AB,  OZ)],  and  the  square  on  AB 
by  the  symbol  sq.  AB.  Let  the  symbol  o  stand  for  the  phrase  "is 
equivalent  to  "  ;  the  sign  +  for  "  added  to  "  or  "  increased  by  "  ;  and 
the  sign  —  for  "diminished  by." 

Since  sq.  AB  =0=  [AB,  AE]  +  [AB,  EB]  ;  [40  (6) 

and  [AB,  AE]^ sq.  AE  +  [AE,  EB],  [40  (a) 

[AB,  EB]^sq.  EB  +  [AE,  EB]  ; 

hence  sq.  AB ^sq.  AE  +  sq.  EB  +  2  [AE,  EB], 

Square  on  sum, 

43.  Cor.  I .  The  square  on  the  sum  of  two  lines  is  equivalent 
to  the  sum  of  their  squares  and  twice  their  rectangle. 

44.  Cor.  2.  Tlie  square  on  any  line  is  equivalent  to  four 
times  the  sqxiare  on  its  half. 


140  PLANE  GEOMETRY  —  BOOK  11 

Ex.  1.  Prove  43  by  applying  the  distributive  property  to  two  lines 
each  equal  to  the  sura  of  the  two  given  lines. 

Ex.  2.  Prove  44  by  applying  the  distributive  property  to  two  equal 
lines  each  of  which  is  bisected. 

Ex.  3.  If  a  line  is  divided  into  three  parts,  the  square  on  the  whole 
line  is  equivalent  to  the  sum  of  the  squares  on  the  parts  together  with 
twice  the  rectangles  of  the  parts  taken  two  and  two. 

Sum  of  squares  on  whole  and  part, 

45.  Theorem  13.  If  a  line  is  divided  into  any  two 
parts,  the  sum  of  the  squares  on  tlve  whole  line  and 
one  part  is  equivalent  to  twice  the  rectangle  of  the 
whole  line  and  that  part,  together  with  the  square  on 
the  other  part. 

Let  ABhQ  the  given  line  divided  at  E. 

To  prove  that  the  sum  of  the  squares  of  J  5  and  EB  is 
equivalent  to  twice  the  rectangle  of  ^i?  and  EBj  together 
with  the  square  on  AE. 

On  AB  describe  a  square,  and  complete  the  construction 
as  in  the  figure  of  the  preceding  theorem. 

The  square  DB  is  equivalent  to  the  sum  of  the  square  DF 
and  the  rectangles  HE  and  GB.  Add  to  each  of  these  equiv- 
alents the  square  FB.  Then  the  sum  of  the  squares  DB  and 
FB  is  equivalent  to  the  sum  of  the  square  DF  and  the  rec- 
tangles HB  and  GB.  Now  the  latter  rectangles  are  each 
equal  to  the  rectangle  oi  AB  and  EB.  Hence  the  theorem 
is  proved. 

Otherwise : 

sq.  AB<>  [AB,  EB]  -{-  \_AB,  AE],      .  [40  (6) 

=0=  lAB,  EB]  +  sq.  AE  +  ^AE,  EB].  [40  (a) 

Add  the  square  on  EB,  then 
sq.  AB  +  sq.  EB<^  \_AB,  EB]  +  sq.  AE -V  \_AE,  EB]  +  sq.  EB, 

^  [AB,  EB]  +  sq.  AE  +  \,AB,  EB],  [40  (a) 

=^sq.  AE+2iAB,EB]. 


EQUIVALENCES  INVOLVING  RECTANGLES      141 

Square  on  difference. 

46.  Cor.  Tlie  square  on  the  difference  of  two  lines  is 
equivalent  to  the  sum  of  their  squares  diminished  by  twice  their 
rectangle. 

Square  on  sum,  of  whole  and  part, 

47.  Theorem  14.  If  a  line  is  divided  into  any  two 
parts,  the  square  on  the  sum  of  the  whole  line  and  one 
part  is  equivalent  to  four  times  the  rectangle  of  the 
whole  line  and  that  part,  together  with  the  square  on 
the  other  part. 

Let  the  line  ^i?  be  divided  at  E. 


To  prove  that  the  square  on  the  sum  of  AB  and  EB  is 
equivalent  to  four  times  the  rectangle  of  AB  and  EB^  together 
with  the  square  on  AE. 

Since  sq.  {AB  -f-  EB)  o=  sq.  AB  +  sq.  EB  +2  [AB,  EB^ ;       [43 

and  sq.  AB  +  sq.  EB  =o  2  [AB,  EB^  -f  sq.  AE ;  [45 

hence  sq.  (AB  +  EB)  =c=  4  [AB,  EB^  4-  sq.  AE. 

48.  Cor.  TJie  square  on  the  sum  of  two  segments  exceeds 
the  square  on  their  difference  by  four  times  their  rectangle. 

Rectangles  of  equal  parts  and  of  unequal  parts, 

49.  Theorem  15.  //  a  line  is  divided  into  two  equal 
parts,  and  also  into  two  unequal  parts,  the  rectangle 
of  the  unequal  parts,  together  with  the  square  on  the 
intermediate  part,  is  equivalent  to  the  square  on  half 
the  line. 

Let  the  line  AB  be  divided  into  equal  parts  at  C,  and  into 
unequal  parts  at  D. 


142 


PLANE  GEOMETRY — BOOK  II 
F  M  E 


To  prove  that  the  rectangle  oi  AD  and  DB  together  with 
the  square  on  CD  is  equivalent  to  the  square  on  CB. 

On  CB  describe  the  square  CBEF.  Through  D  draw  DM 
perpendicular  to  CB  and  meeting  the  diagonal  BF  in  H. 
Through  H  draw  the  line  GHKL  parallel  U)  AB  \  and  com- 
plete the  rectangle  ACKL. 

The  figures  DG,  KM  are  squares;  and  the  rectangles  CH, 
HE  are  equivalent  (36). 

Add  to  each  of  these  the  figure  DG ;  then  the  rectangles 
CG  and  DE  are  equivalent. 

Now  the  rectangles  AK  and  CG  are  equivalent,  because  AC 
equals  CB  and  CK  is  common. 

Therefore  the  rectangles  AK  and  DE  are  equivalent. 

Add  the  rectangle  CH  and  also  the  square  KM ;  then  the 
rectangle  AH  and  the  square  KM  are  together  equivalent  to 
the  square  CE. 

Now  AH  is  the  rectangle  of  AD  and  DH,  that  is  of  AD  and 
DB ;  and  the  square  KM  is  equal  to  the  square  on  CD. 

Therefore  the  rectangle  oi  AD  and  DB  together  with  the 
square  on  CD  is  equivalent  to  the  square  on  CB. 

Otherwise  : 

[AD,  DB^  =c=  [AC,  DB^  +  [CD,  DB^.  [39 

Replace  AC\)y  its  equal  CB^  and  add  the  square  on  CD,  then 
[AD,  Z>5]  +  sq.  CD^ICB,  DB]  +  [CD,  DB]  +  sq.  CD, 

<^[CB.  DB]  +  [CB,  CD],  [40  (a) 

<>sq.CB.  [40  (ft) 


EQUIVALENCES  INVOLVING  RECTANGLES     143 

50.  Cor.  I.  The  rectangle  of  any  two  lines,  together  with 
the  square  on  half  their  differerice,  is  equivalent  to  the  square  on 
half  their  sum. 


A  P      i       D  B 

Let  AD  and  DB  be  the  lines,  AB  their  sum  and  CB  their 
half  sum. 

Take  CD'  equal  to  CD.  Then  DB  equals  AD',  and  the 
difference  of  the  two  lines  AD  and  DB  is  equal  to  the  differ- 
ence of  AD  and  AD',  which  is  D'd. 

Therefore  CD  is  half  the  difference  of  AD  and  DB. 

Now  the  rectangle  of  AD  and  DB  together  with  the  square 
on  CD  is  equivalent  to  the  square  on  CB  (49). 

Hence  the  rectangle  of  two  lines,  together  with  the  square 
on  half  their  difference,  is  equivalent  to  the  square  on  half 
their  sum. 


Difference  of  two  squares  expressed  as  a  rectangle. 

51.  Cor.  2.  The  rectangle  of  the  sum  and  difference  of  tivo 
lines  is  equivalent  to  the  difference  of  the  squares  on  the  lines. 

Let  AC,  CD  be  the  two  lines;  then  AD  is  their  sum,  and 
DB  is  their  difference. 

Hence,  by  49,  the  rectangle  of  the  sum  and  difference  of 
two  lines,  together  with  the  square  on  the  less,  is  equivalent 
to  the  square  on  the  greater. 

In  other  words,  the  rectangle  of  the  sum  and  difference  of 
two  lines  -is  equivalent  to  the  difference  of  the  squares  on 
the  lines. 

Ex.  1.  If  there  are  two  given  squares,  show  how  to  construct  a 
rectangle  equivalent  to  their  difference. 

Ex.  2.  If  a  line  is  divided  into  two  equal  parts  and  also  into  two 
unequal  parts,  show  that  the  rectangle  of  the  unequal  parts  is  less  than 
the  rectangle  of  the  equal  parts. 


144  PLANE  GEOMETRY  —  BOOK  II 

Modification  of  49, 

52.  Theorem  16.  //  a  given  line  is  bisected  and  then 
extended  to  any  point,  the  rectangle  contained  by  the 
extension  and  the  whole  line  so  extended,  together  with 
the  square  on  half  the  original  line,  is  equivalent  to 
the  square  on  the  line  between  the  point  of  bisection 
and  the  point  of  extension. 


A  c 

Prove  as  in  theorem  preceding. 

Otherwise : 

[AD,  BD] o [AC,  BD]  +  [CZ),  BD].  [39 

Replace  AG  by  its  equal  CB,  and  add  the  square  on  CB  ;  then 
IAD,  BD]  +  sq.  CB^[CB,  BD]  +  sq.  CB  +  [CD,  BD], 

o  [CA  CB]  +  [CD,  BD],  [40  (a) 

=c=sq.  CD.  [40  (6) 

53.  Cor.  Show  that  CB  is  half  the  difference  of  AD,  DB  ; 
and  that  CD  is  half  the  sum  of  AD,  DB ;  and  hence  prove  again 
that  "  the  rectangle  of  two  lines  together  with  the  square  on  half 
their  difference  is  equivalent  to  the  square  on  half  their  sum.'' 

Ex.  Prove  again  that  "  the  rectangle  of  the  sum  and  difference  of 
two  lines  is  equivalent  to  the  difference  of  their  squares." 

[Let  CD,  CB  be  the  segments,  AD  their  sum,  BD  their  difference.] 


EQUIVALENCES  INVOLVING  RECTANGLES     145 

Squares  on  equal  parts  and  on  unequal  parts, 

54.  Theorem  17.  //  a  line  is  divided  into  two  eqiud 
parts  and  also  into  unequal  parts,  the  sum  of  tlie 
squares  on  the  unequal  parts  is  equivalent  to  double 
the  sum  on  the  squares  on  the  half  line  and  on  the 
intermediate  part. 

Let  AB  be  the  given  line,  divided  into  two  equal  parts  at 
C  and  into  two  unequal  parts  at  D. 


A  i  D  B 

To  prove  that  the  sum  of  the  squares  on  AD,  DB  is  equiva- 
lent to  double  the  sum  of  the  squares  on  J.C,  CD. 

By  43  and  46 

sq.  AD^sq.  AC  +  sq.  CD  +  2  [AC,  CD], 
sq.  DB  =0=  sq.  CB  +  sq.  CD  -  2  [CD,  CD^. 

Add  these  equivalents,  observing  that  AC  equals  CD,  and 
that  the  equal  rectangles  disappear  since  one  is  added  and 
the  other  subtracted.     Therefore 

sq.  AD  +  sq.  DD  =0=  2  sq.  AC +  2  sq.  CD. 

55.  Cor.  I.  The  sum  of  the  squares  on  any  two  lines  is 
equivaleyit  to  twice  the  square  on  half  their  sum  together  with 
twice  the  square  on  half  their  difference.     (See  50.) 

56.  Cor.  2.  The  square  on  the  sum  of  tivo  lines  together 
with  the  square  on  their  difference  is  equivalent  to  double  the 
sum  of  the  squares  on  the  two  lines. 

[Let  AC,  CD  be  the  given  lines,  AD  their  ^um,  DB  their  differ- 
ence.] 

Ex.    Prove  56  directly  from  43  and  46. 


146  PLANE  GEOMETRY— ROOK  II 

Modification  of  54, 

57.  Theorem  18.  If  a  given  line  is  bisected  and  then 
extended  to  any  point,  the  sum  of  the  squares  on  tJie 
extension  and  on  tJie  whole  line  so  extended  is  equivor 
lent  to  twice  the  square  on  half  the  original  line, 
together  with  twice  the  square  on  the  line  between  the 
point  of  bisection  and  the  point  of  extension. 

Let  AB  be  the  given  line,  bisected  at  C,  and  then  extended 
toD. 

i  Z  ;§      ~D 

To  prove  that 

sq.  AB  +  sq.  52)  =c=  2  sq.  ^C  4-  2  sq.  CD, 

Show  that  the  proof  of  the  preceding  theorem  applies,  letter  by 
letter,  to  this  theorem. 

Cotnbined  statement  of  theorems  17 ^  18, 

In  order  to  combine  these  two  theorems  in  one  statement 
an  extended  meaning  will  now  be  given  to  the  phrase  "  the 
two  segments  of  a  line." 

58.  Definition.  If  on  the  line  AB  the  point  C  is  taken 
between  A  and  B,  then  the  line  AB  is  said  to  be  divided 
internally  into  the  two  segments  AC,  BC. 

Again,  if  the  point  C  is  taken  on  the  prolongation  of  AB, 
then  the  line  AB  i^  said  to  be  divided  externally  into  the 
two  segments  AC,  BC. 

59.  Restatement.  The  two  theorems  may  then  be  re- 
stated as  follows : 

If  a  given  line  is  bisected  and  divided  unequally 
{either  internally  or  externally),  then  the  sum  of  the 
squares  on  the  unequal  parts  is  equivalent  to  twice  the 
square  on  half  the  original  line,  together  with  twice 
the  square  on  the  line  between  the  points  of  division. 


EQUIVALENCES  IN  A    TRIANGLE 


147 


EXERCISES 

1.  The  square  on  the  sum  of  two  lines  is  greater  than  the  sum  of 

the  squares  on  the  two  lines. 

2.  The  sum  of  the  squares  on  two  lines  is  never  less  than  twice  their 
rectangle. 

3.  If  a  line  is  divided  into  two  equal  parts  and  also  into  two  unequal 
parts,  how  does  the  sura  of  the  squares  on  the  equal  parts  compare 
with  the  sum  of  the  squares  on  the  unequal  parts  ? 

4.  If  a  line  is  divided  into  two  equal  parts  and  also  into  two  unequal 
parts,  then  the  sum  of  the  squares  on  the  unequal  parts  exceeds  twice 
their  rectangle  by  four  times  the  square  on  the  intermediate  segment. 


EQUIVALENCES   IN  A  TRIANGLE 

Relations  in  a  right  triangle. 

60.  Theorem  19.  In  a  right  triangle  the  rectangle 
of  the  hypotenuse  and  the  projection  upon  it  of  one 
of  the  other  sides  is  equivalent  to  the  square  on  that 
side. 

Let  the  triangle  ABC  have  the  angle  C  a  right  angle,  and 
let  BD  he  the  projection  of  the  side  BG  upon  the  hypote- 
nuse BA. 

K 


To  prove  that  the  rectangle  of  ^^  and  DB  is  equivalent 
to  the  square  on  BC. 

On  BC  describe  the  square  BCFE.  Prolong  EF  to  meet 
in  H  the  line  BH  drawn  perpendicular  to  AB.  Complete 
the  rectangle  DBHK,  and  join  CH. 


148  PLANE  GEOMETRY  —  BOOK  II 

In  the  triangles  BCA  and  BEU,  the  angles  ABC  and  EBH 
are  equal,  being  each  complemental  to  CBH;  also  the  angles 
BCA  and  BEH  Sive  equal,  being  right;  and  the  sides  BC  and 
BE  are  equal,  being  sides  of  a  square.  Hence  the  side  BA 
equals  BH  (I.  65). 

Therefore  the  rectangle  BK  is  the  rectangle  of  AB  aud 
DB. 

Now  this  rectangle  is  double  the  triangle  CBHy  since  they 
have  the  same  base  BHy  and  the  same  altitude  DB  (29). 

Also  the  square  BF  is  double  the  same  triangle,  since 
they  have  the  same  base  BC  and  the  same  altitude  BE. 

Therefore  the  rectangle  and  square  are.  equivalent  (9). 

That  is  to  say,  the  square  on  the  side  BC  is  equivalent  to 
the  rectangle  of  the  projection  BD  and  the  hypotenuse  BA. 

In  the  same  way  it  can  be  proved  that  the  square  on  the 
side  JC  is  equivalent  to  the  rectangle  of  its  projection  AD 
and  the  hypotenuse  BA. 

Theoretn  of  Pythagoras. 

61.  Theorem  20.  In  any  right  triangle  the  square 
on  the  hypotenuse  is  equivalent  to  the  sum  of  the 
squares  on  the  other  two  sides. 

Let  ^5C  be  a  triangle  having  c  a  right  angle. 

To  prove  that  the  square  on  the  hypotenuse  AB  is 
equivalent  to  the  sum  of  the  squares  on  the  sides  ACy  CB. 

The  square  on  AB  is  equivalent  to 
the  sum  of  the  rectangles  oi  AB  and 
ADy  and  oi  AB  and  BD  [40  (cor.  1)]. 

Now  the  rectangle  of  ^5  and  AD 
is  equivalent   to  the   square   on  ^C 
(60);   and  the  rectangle  of  AB  and  DB  is   equivalent  to 
the  square  on  CB. 

Therefore  the  square  on  AB  is  equivalent  to  the  sum  of 
the  squares  on  AC  and  CB, 


EQUIVALENCES  IN  A   TRIANGLE 


149 


Ex.    Show  how  to  dissect  the  squares  on  ^Cand  CB  so  that  the 
parts  may  cover  the  square  on  AB. 


\ 


Note.  The  earliest  proof  of  this  cele- 
brated theorem  is  attributed  to  Pythagoras 
(550  B.C.),  the  founder  of  the  famous  Pytha- 
gorean School  in  lower  Italy.  The  theorem 
itself  was,  however,  probably  known  as  an 
experimental  fact  to  the  ancient  Egyptians, 
a  thousand  years  earlier.  It  is  conjectured 
that  the  Pythagorean  proof  was  based  on 
some  method  of  dissection  similar  to  that 
shown.  The  proof  given  by  Euclid  (300  b.c.)  is  a  combination  of  60 
and  61.  The  first  part  is  here  enunciated  as  a  separate  theorem  on 
account  of  its  great  importance. 


Relation  in  an  obtuse  triangle. 

62.  Theorem  21.  In  an  obtuse-angled  triangle  the 
square  on  the  side  opposite  the  obtuse  angle  is  greater 
than  the  sum  of  the  squares  on  the  other  two  sides 
by  twice  the  rectangle  contained  by  either  of  these 
sides  and  the  projection  of  the  other  upon  it. 

Let  the  triangle  ABC  have  the  angle  C  obtuse,  and  let  CD 
be  the  projection  of  the  side  CB  on  AC  extended. 

To  prove  that  the  square  on  ^J5  is 
greater  than  the  sum  of  the  squares 
on  ^C  and  CB  by  twice  the  rectangle 
of  ^C  and  CD. 

The  square  on  4Z>  is  equivalent  to 
the  sum  of  the  squares  on  AC  and 
CD  J  together  with  twice  the  rectangle 
of  ^C  and  CD  (42). 

Add  to  each  member  of  this  equivalence  the  square  on  BD. 

Then  the  sum  of  the  squares  on  AD  and  BD  is  equivalent 
to  the  sum  of  the  squares  on  AC,  CD,  and  DB  together  with 
twice  the  rectangle  of  ^C7  and  CD. 

Now  the  sum  of  the  squares  on  AD  and  DB  is  equivalent 


150  PLANE  GEOMETRY  —  BOOK  II 

to  the  square  on  AB  (61);  and  the  sum  of  the  squares  on 
CD  and  DB  is  equivalent  to  the  square  on  CB. 

Therefore  the  square  on  AB  is  equivalent  to  the  sum  of 
the  squares  on  ^C  and  C5,  together  with  twice  the  rectangle 
of  ^C  and  CD. 

In  other  words,  the  square  on  AB  exceeds  the  sum  of  the 
squares  on  AC  and  BC  hy  twice  the  rectangle  of  ^C  and  CD. 

Relations  in  any  triangle. 

63.  Theorem  22.  In  any  triangle  the  square  on  the 
side  opposite  an  a^icte  angle  is  less  than  the  sum  of 
the  squares  on  the  sides  containing  that  angle  hy 
twice  the  rectangle  of  either  of  these  sides  and  the 
projection  of  the  other  upon  it. 

Let  ABC  be  a  triangle  having  the  angle  C  acute,  and  let 
DC  be  the  projection  of  the  side  BC  upon  the  side  AC. 

To  prove  that  the  square  on  AB  is 
less  than  the  sum  of  the  squares  on 
AC  and  CB  by  twice  the  rectangle 
of  AG  and  DC. 

The  sum  of  the  squares  on  ^c  and 
DC  is  equivalent  to  twice  the  rectangle 
of  ^C  and  DC  together  with  the  square 
on  AD  (45). 

To  each  member  of  this  equivalence  add  the  square  on  DB. 

Then  the  sum  of  the  squares  on  AC,  CD,  DB  is  equivalent 
to  twice  the  rectangle  of  AC  and  DC  together  with  the  sum 
of  the  squares  on  AD  and  DB. 

But  the  sum  of  the  squares  on  DC  and  DD  is  equivalent 
to  the  square  on  BC  (61) ;  and  the  sum  of  the  squares  on 
AD,  DB  is  equivalent  to  the  square  on  AB. 

Therefore,  the  sum  of  the  squares  on  ^  C  and  CB  is  equiva- 
lent to  twice  the  rectangle  of  ^C  and  DC  together  with  the 
square  on  AB. 

That  is,  the  square  on  AB  \^  less  than  the  sum  of  the 
squares  on  AC  and  CB  by  twice  the  rectangle  of  ^C  and  DC. 


EQUIVALENCES  IN  A    TRIANGLE  151 

64.  Cor.  I.  In  any  triangle,  according  as  one  angle  is 
greater  than,  equal  to,  or  less  than  a  right  angle,  so  is  the 
square  on  the  opposite  side  greater  than,  equivalent  to,  or  less 
than  the  sum  of  the  squares  on  the  other  two  sides.  (61,  62, 63.) 

65.  Cor.  2.  In  any  triangle,  according  as  the  square  on  one 
side  is  equivalent  to,  greater  than,  or  less  than  the  sum  of  the 
squares  on  the  other  tivo  sides,  so  is  the  angle  opposite  the  first 
side  greater  than,  equal  to,  or  less  than  a  right  angle. 

Melatlon  involving  perpendicular. 

66.  Theorem  23.  In  any  triangle,  if  a  perpendic- 
ular is  dramn  from  the  vertex  to  the  hase,  then 
according  as  the  vertical  angle  is  greater  than,  equal 
to,  or  less  than  a  right  angle,  so  is  the  rectangle  of 
the  segments  of  the  base  greater  than,  equivalent  to, 
or  less  than  the  square  on  the  perpendicular. 

Let  BD  he  the  perpendicular  from  the  vertex  to  the  base 
in  the  triangle  ABC. 

To  prove  that  the  square  on  BD  is  equivalent  to,  greater 
than,  or  less  than  the  rectangle  of  AD  and  DC  according  as 
the  angle  B  is  right,  acute,  or  obtuse. 

The  square  on  AC  is  equivalent  to,  less  than,  or  greater 
than  the  sum  of  the  squares  on  AB  and  CB  according  as  the 
angle  B  is  right,  acute,  or  obtuse  (64). 

Now  the  square  on  AG  is  equivalent  to  the  sum  of  the 
squares  on  AD  and  DC  with  twice  the  rectangle  of  ^/>*and 
DC  (42) ;  the  square  on  AB  is  equivalent  to  the  sum  of  the 
squares  on  AD  and  DB ;  and  the  square  on  BC  is  equivalent 
to  the  sum  of  the  squares  on  DC  and  DB  (61). 

Rejecting  the  common  sum  of  squares  on  AD  and  DC,  it 
remains  that  twice  the  rectangle  oi  AD  and  DC  is  equivalent 
to,  less  than,  or  greater  than  twice  the  square  on  DB, 
according  as  the  angle  B  is  right,  acute,  or  obtuse ;  whence 
by  taking  halves  the  theorem  follows  (11,  12). 

MCM.  ELEM.  GEOM.  —  11 


152 


PLANE  GEOMETRY  —  BOOK  II 


Relation  involving  median, 

67.  Theorem  24.  In  any  triangle,  if  a  line  is 
drawn  from  the  vertex  to  the  mid-point  of  the  base, 
the  sum  of  the  squares  on  the  two  sides  is  equiva- 
lent to  twice  the  square  on  half  the  ba^e,  together 
with  twice  the  square  on  the  median  line. 

Let  BD  be  a  median  line  of  the  triangle  ABC. 

To  prove  that  the  sum  of 
the  squares  on  the  sides  AB 
and  5C  is  equivalent  to  twice 
the  square  on  DC  and  twice 
the  square  on  BD. 

Case  1.  Let  the  angles  BDA 
and  BDC  be  equal. 

Then  BD  is  perpendicular  to  AC.  Therefore  the  square 
on  AB  is  equivalent  to  the  sum  of  the  squares  on  AD  and 
BD  (61) ;  and  the  square  on  BC  is  equivalent  to  the  sum  of 
the  squares  on  BD  and  DC. 

Hence  the  sum  of  the  squares  on  AB  and  BC  is  equivalent 
to  twice  the  square  on  BDj  together  with  twice  the  square 
on  AD. 

Case  2.   Let  the  angles  BDA  and  BDC  be  unequal. 

Then  one  of  them  is  the  greater.  Let  the  angle  ADB  be 
greater  than  BDC.  The  angle  ADB  is  then  obtuse,  and  BDC 
is  acute. 

Let  BE  be  the  perpendicular  from  B  to  the  base  AC. 

Then  in  the  triangle  ADB  the  square  on  AB  is  equivalent 
to  the  sum  of  the  squares  on  AD  and  BD  together  with  twice 
the  rectangle  of  AD  and  DE  (62). 

Again,  in  the  triangle  BDC,  the  square  on  BC  is  equivalent 
to  the  sum  of  the  squares  on  BD  and  DC  diminished  by 
twice  the  rectangle  oi  DC  and  DE  (63). 

Add  these  equivalences,  member  to  member,  and  reject 
the  equivalent  double  rectangles. 


EQUIVALENCES  IN  A   THIANGLE  153 

Then  the  sum  of  the  squares  on  AB  and  BC  is  equivalent 
to  twice  the  square  on  BD  and  twice  the  square  on  AD. 

68.  Cor.  The  sum  of  the  squares  on  the  four  sides  of  a 
parallelogram  is  equivalent  to  the  sum  of  the  squares  on  the 
two  diagonals.     (Use  I.  159 ;  and  67,  44.) 

Ex.  The  sum  of  the  squares  on  the  four  sides  of  a  quadrilateral  is 
equivalent  to  the  sum  of  the  squares  on  the  two  diagonals,  and  four 
times  the  square  on  the  segment  joining  the  mid-points  of  the  diagonals. 

Difference  of  squares  on  sides, 

69.  Theorem  25.  In  any  triangle,  if  a  line  is 
drawn  from  the  vertex  perpendicular  to  the  base 
{or  base  prolonged),  then  the  difference  of  the  squares 
on  the  two  sides  is  equivalent  to  the  difference  of 
the  squares  on  the  segments  of  the  base. 

[Take  the  figure  of  62,  or  63.  Apply  61  to  each  of  the  right 
triangles.     Then  subtract.] 

70.  Cor.  In  an  isosceles  triangle,  if  a  line  is  drawn  from 
the  vertex  to  any  point  of  the  base  (or  base  prolonged),  then  the 
difference  of  the  squares  on  this  line  and  on  one  side  of  the 
triangle  is  equivalent  to  the  rectangle  of  the  segments  of  the  base. 

Outline.  Let  BA,  BC  be  the  equal  sides,  P  any  point  on  the  base 
AC.  Draw  perpendicular  JSil/,  which  bisects  ^C  at  Jf.  Apply  69  to 
the  triangle  ABP.    Then  use  50.    Give  the  proof  in  the  usual  form. 

EXERCISES 

1.  In  an  isosceles  right  triangle  the  square  on  one  side  is  equivalent 
to  half  the  square  on  the  hypotenuse ;  and  the  square  on  the  perpen- 
dicular is  equivalentlo  one  fourth  of  the  square  on  the  hypotenuse. 

2.  If  the  acute  angles  of  a  right  triangle  are  respectively  equal  to 
one  third  and  two  thirds  of  a  right  angle,  then  the  squares  on  the 
opposite  sides  are  respectively  equivalent  to  one  fourth  and  three 
fourths  of  the  square  on  the  hypotenuse. 

3.  In  the  same  case  the  square  on  the  perpendicular  is  equivalent 
to  three  fourths  of  the  square  on  the  side. 


154  PLANE  GEOMETRY  —  BOOK  II 

4.  In  an  equilateral  triangle  the  square  on  the  altitude  is  equivalent 
to  three  fourths  of  the  square  on  the  side. 

5.  In  an  isosceles  triangle,  if  a  perpendicular  is  drawn  from  an 
extremity  of  the  base  to  the  opposite  side,  then  twice  the  rectangle 
contained  by  that  side  and  its  segment  adjacent  to  the  base  is  equiva- 
lent to  the  square  on  the  base. 

6.  In  any  triangle,  if  an  angle  is  equal  to  two  thirds  of  a  straight 
angle,  then  the  square  on  the  side  opposite  is  equivalent  to  the  sum  of 
the  squares  on  the  other  two  sides  and  the  rectangle  contained  by  them. 

7.  If  any  point  is  joined  to  the  four  vertices  of  a  rectangle,  the  sum 
of  the  squares  on  the  lines  drawn  to  two  opposite  vertices  is  equivalent 
to  the  sum  of  the  squares  on  the  other  two  joining  lines. 

CONSTRUCTION  OF  EQUIVALENT  POLYGONS 

Reduction  of  polygon  to  equivalent  triangle, 

71.  Problem  1.  To  construct  a  triangle  equivalent 
to  a  given  polygon. 

Let  ABODE  be  the  given  polygon. 

To  construct  a  triangle  equivalent 
to  it. 

Draw  any  diagonal  AC  connect- 
ing the  ends  of  two  adjacent  sides. 
Through  the  intermediate  vertex  B 
draw  BH  parallel  to  this  diagonal  to 

meet  one  of  the  sides  next  in  order,  say  EA,  in  Hy  and 
draw  CH. 

The  triangles  CAB  and  CAH  are  equivalent  (33). 

To  each  add  the  polygon  ACDE-^  then  the  given  polygon 
ABODE  is  equivalent  to  the  polygon  EHOD. 

The  number  of  sides  of  the  latter  polygon  is  one  less  than 
the  number  of  sides  of  the  given  polygon. 

Repeating  this  process  a  set  of  equivalent  polygons  having 
fewer  and  fewer  sides  is  obtained;  and  the  process  ends 
when  a  three-sided  figure  is  reached. 


CONSTRUCTION  OF  EQUIVALENT  POLYGONS      155 
Conversion  of  triangle  into  equivalent  rectangle, 

72.  Problem  2.  To  construct  a  parallelogram  equiva- 
lent to  a  given  triangle,  and  having  an  angle  equal  to 
a  given  angle. 

Let  ABChQ  the  given  triangle,  and  M  the  given  angle. 


To  construct  a  parallelogram  equivalent  to  the  triangle 
ABC,  and  having  an  angle  equal  to  M. 

Bisect  AB  at  D,  and  draw  DE,  making  the  angle  BDE  equal 
to  M.  Draw  CEF  parallel  to  AB,  and  complete  the  parallelo- 
gram DBFE. 

This  parallelogram  is  equivalent  to  the  triangle  ABC. 

To  prove  this,  draw  CD. 

The  parallelogram  DBFE  is  double  the  triangle  BDC,  since 
they  have  the  same  base  and  equal  altitudes  (34). 

The  triangles  ADC,  DBC  are  equivalent,  having  equal  bases 
and  the  same  altitude  (29).  Hence  the  triangle  ABC  is 
double  the  triangle  DBC. 

Therefore  the  parallelogram  DBFE  is  equivalent  to  the 
given  triangle  ABC  (9);  and  it  has  an  angle  equal  to  the 
given  angle. 

Ex.  1.  To  construct  a  rectangle  that  shall  be  equivalent  to  a  given 
triangle. 

Ex.  2.  To  construct  an  isosceles  triangle  equivalent  to  a  given 
triangle. 

Ex.  3.  To  construct  a  parallelogram  equivalent  to  a  given  parallel- 
ogram, and  having  an  angle  equal  to  a  given  angle. 


156  PLANE  GEOMETRY — BOOK  II 

Conversion  of  parcUlelograin  into  equivalent  one, 

73.  Problem  3.  On  a  given  line  to  CA)nstruct  a 
parallelogram  equLivalent  to  a  given  parallelogram, 
and  having  its  angles  equal  to  tJie  angles  of  this 
parallelogram. 

Let  ABCD  he  the  given  parallelogram  and  FK  the  given 
line. 

D c  HO  L 

I — 7      r^T      ":-~^ 

LJ       eU,J^ /k 


It  is  required  to  construct  on  FK  a  parallelogram  equivalent 
to  ABCD  and  having  its  angles  respectively  equal  to  the 
angles  of  ABCD. 

Prolong  KF,  and  lay  off  FE  equal  to  BA^  one  of  the  sides 
of  the  given  parallelogram. 

Transfer  the  figure  ABCD  to  the  position  EFGH,  so  that 
BA  may  fall  on  FE  (I.  199). 

Draw  KL  parallel  to  FG  to  meet  IIG  extended  in  L.  Draw 
LF^  and  prolong  it  to  meet  HE  extended  in  M.  Draw  MP 
parallel  to  EF,  and  let  it  meet  the  extensions  of  GF  and  LK 
in  N  and  P. 

Then  NPKF  is  the  required  parallelogram. 

For  it  is  equivalent  to  EFGH  (36) ;  its  angles  are  equal 
to  the  angles  of  EFGH  (I.  126) ;  and  it  is  described  on  the 
given  line  FK. 

73  (a).  Cor.  On  a  given  line  to  construct  a  rectangle  equiva- 
lent to  a  given  rectangle. 

Ex,  1.  Given  one  side  of  a  rectangle  and  the  equivalent  square, 
find  the  adjacent  side. 

Ex.  2.  On  a  given  line  to  construct  a  rectangle  equivalent  to  a  given 
triangle  (72,  73). 


CONSTRUCTION  OF  EQUIVALENT  POLYGONS      157 
To  ** square''  a  rectangle, 

74.   Problem  4.     To  construct  a  square  equivalent 
to  a  given  rectangle. 

Let  ABCB  be  the  given  rectangle. 


/ 

t 
A,— 


H 


D 


B       E 


To  construct  a  square  equivalent  to  it. 

Prolong  AB  to  E,  making  BE  equal  to  BC.  Bisect  AE  at 
H  (I.  70).  With  H  as  center  and  HE  as  radius  describe  the 
arc  ERA.     Prolong  CB  to  meet  this  arc  in  K. 

The  square  constructed  on  BK  is  equivalent  to  the  given 
rectangle. 

To  prove  this,  draw  the  radius  HK. 

The  rectangle  oi  AB  and  BE  with  the  square  on  HB  is 
equivalent  to  the  square  on  HE  (49) ;  that  is,  to  the  square 
on  HK,  which  is  equivalent  to  the  sum  of  the  squares  on  HB 
and  BK  (61). 

Reject  the  common  square  on  HB.  Then  the  rectangle  of 
AB  and  BE  is  equivalent  to  the  square  on  BK. 

Now  the  rectangle  ABCD  is  the  rectangle  of  AB  and  BC, 
that  is,  oi  AB  and  BE. 

Therefore  the  square  on  BK  is  equivalent  to  the  given 
rectangle. 

75.  Summary.  By  combining  the  constructions  given  in 
problems  1,  2,  4,  a  square  can  be  constructed  equivalent  to 
any  given  polygon. 

This  square  is  called  its  equivalent  square;  and  the 
process  of  construction  is  called  squaring  the  polygon. 


158  PLANE  GEOMETRY  —  BOOK  II 

Ex.  1.    Give  the  complete  construction  for  squaring  a  given  triangle. 

Ex.  2.    Also  for  squaring  a  given  quadrangle. 

Ex.  3.  If  the  lines  AK,  KE  are  drawn,  prove  that  the  angle  AKE 
is  equal  to  the  sum  of  the  angles  KAE^  AEK\  and  hence  that  AKE 
is  a  right  angle. 

Ex.  4.  Use  ex.  3  to  construct  on  a  given  line  a  rectangle  equivalent 
to  a  given  square. 

[Here  AB,  BKare  given,  to  find  BE.    Compare  with  73,  ex.  1.] 

Addition  of  squares, 

76.  Problem  5.  To  construct  a  square  equivalent 
to  the  sum  of  two  given  squares. 

Let  KLj  MN  be  the  sides  of  the  given  squares. 

n 

M iV 

K  L 


To  construct  a  square  equivalent  to  the  sum  of  the  squares 
on  these  lines. 

Draw  CA  equal  to  KL.  Erect  CB  perpendicular  to  CA 
and  equal  to  MN.     Draw  the  line  AB. 

The  square  on  AB  is  equivalent  to  the  sum  of  the  squares 
on  KL  and  MN. 

For  the  square  on  AB  is  equivalent  to  the  sum  of  the 
squares  on  AC  and  CB  (61)  and  hence  equivalent  to  the  sum 
of  the  squares  on  KL  and  MN. 

Ex.  To  constnict  a  square  equivalent  to  the  sum  of  three  or  more 
given  squares. 

Subtraction  of  squares, 

77.  Problem  6.  To  construct  a  square  equivalent 
to  the  difference  of  two  given  squares. 

[This  is  a  particular  case  of  I.  136.] 

Ex.  Show  how  to  construct  a  square  equivalent  to  the  difference  of 
two  polygons. 


DIVISION  OF  A   LINE  159 

To  halve  a  square, 

78.   Problem  7.     To  construct  a  square  equivaUnt 
to  tialf  a  given  square. 

Let  ABhQ  the  side  of  the  given  square. 
To  construct  a  square  equivalent 
to  half  the  square  on  AB. 

Draw  AC,  BC,  making  the  angles 
BAC,  ABC  each  equal  to  half  a  right 
angle. 

Show  that  the  square  on  AC  is  half  the  square  on  AB. 

DIVISION  OF  A  LINE 

In  the  following  problems  a  given  line  is  to  be  divided  so 
that  the  two  parts  may  fulfill  given  conditions :  e.g.,  have  a 
given  difference  ;  a  given  sum  of  squares ;  etc. 

lyifference  given, 

79.  Problem  8.  To  divide  a  given  line  into  two 
parts  so  that  the  difference  of  the  parts  shall  be  equal 
to  another  given  line. 

Let  AB,  CD  be  the  given  lines,  of  which  AB  is  the  greater. 

To   divide   AB    into  two 

c  n 

parts  whose  difference  shall 

be  equal  to  CD. 

Prolong  AB,  making  BE    "^  F     B  E 

equal  to  CD.     Bisect  AE  in  F.     This  point  F  divides  AB  so 
that  the  difference  between  AF  and  FB  is  equal  to  CD. 

For,  since  AF  equals  FE,  the  difference  between  AF  and 
FB  is  equal  to  the  difference  between  FE  and  FB,  which  is 
equal  to  be,  that  is  to  CD. 

Bestriction.  The  line  CD  must  be  less  than  AB,  otherwise  there 
is  no  solution  to  the  problem. 


160  PLANE  GEOMETRY  —  BOOK  II 

Difference  of  squares, 

80.  Problem  9.  To  divide  a  given  line  internally 
so  that  the  difference  of  the  squares  on  the  two  seg- 
ments may  he  equivalent  to  a  given  square. 

Let  AB  hQ  the  given  line,  and  let  CD  be  the  side  of  the 
given  square.  ^  P 

To  divide  AB  so  that  the  ^^''^  ' 

difference  of  the  squares  on  ^^''^ 

the  parts  may  be  equivalent       ^.-'^ ^__ 

to  the  square  on  CD.  ^  M       B      L         C 

Erect  BK  perpendicular  to  AB  and  equal  to  CD.  Draw 
AKj  and  erect  KL  perpendicular  to  AKj  meeting  AB  extended 
in  L.     Bisect  AL  in  M. 

Then  M  divides  AB  so  that  the  difference  of  the  squares 
on  ^if  and  MB  is  equivalent  to  the  square  on  CD. 

For  the  rectangle  of  AB  and  BL  is  equivalent  to  the  square 
on  BK  (66).  Also  the  rectangle  of  ^JS  and  BL  with  the 
square  on  MB  is  equivalent  to  the  square  on  ML  (49). 

Hence  the  rectangle  of  ^J5  and  BL  is  equivalent  to  the 
difference  of  the  squares  on  AM  and  MB. 

Therefore  the  difference  of  the  squares  on  AM  and  MB  is 
equivalent  to  the  square  on  BK,  that  is  to  the  square  on  CD. 

'  Restriction.    The  given  square  must  be  less  than  the  square  on  the 
given  line. 

81.  Note.  This  problem  may  also  be  solved  by  drawing  KM  so 
as  to  make  the  angle  ^A'Jf  equal  to  MAK,  and  then  proving  by  61. 

82.  Cor.   To  divide  a  given  y'\ 
line  exteiiially  so  that  the                         y^     !  ^^ 
difference  of  the  squares  on                     y 
the   two   segments  shall   be                y 
equivalent  to  a  given  square,          /" 

nahen  the  latter  is  greater  than    ^ ^ ^y 

the  square  on  the  given  line.      


DIVISION  OF  A  LINE  161 

Rectangle  given* 

83.  Problem  10.  To  divide  a  given  line  internally 
so  that  the  rectangle  of  the  two  segments  may  be 
equivalent  to  a  given  square. 

Let  AB  hQ  the  given  line,  and  HK  the  side  of  the  given 
square. 


H 


To  divide  AB  into  two  parts  AD  and  DB  so  that  the 
rectangle  of  AD  and  DB  may  be  equivalent  to  the  square 
on  HK. 

Bisect  AB  at  M.  With  JWas  center  and  radius  MB  describe 
the  arc  ACB.  Erect  BE  perpendicular  to  AB  and  equal  to 
HK.  Draw  EC  parallel  to  BA,  meeting  the  arc  in  C.  Draw 
CD  perpendicular  to  AB. 

The  point  D  divides  ^2?  so  that  the  rectangle  oi  AD  and 
2)5  is  equivalent  to  the  square  on  HK. 

Prove  as  in  74. 

Bestriction.  The  given  square  must  be  less  than  the  square  on  half 
the  given  line. 

84.  Note.  This  problem  can  also  be  solved  by  the  use  of  61  and 
73,  etc. 

85.  Problem  11.  To  divide  a  given  line  externally 
so  that  the  rectangle  of  tJie  two  segments  shall  he 
equivalent  to  a  given  square. 

Outline.  Let  AB  be  the  given  line ;  draw  the  perpendicular  EC 
equal  to  side  of  given  square;  take  ilf  mid-point  of  AB ;  join  MC\  on 
AB  prolonged  lay  off  MD  equal  to  MC.  Prove  by  52  that  rectangle  of 
AD  and  BD  is  equivalent  to  the  square  on  EC. 


162  PLANE  GEOMETRY — BOOK  II 

Sum  of  squares  given. 

86.  Problem  12.  To  divide  a  given  line  internally 
so  that  the  sum  of  the  squares  on  the  parts  may  be 
equivalent  to  a  given  square. 

Let  AB  be  the  given  line,  and  HK  a  side  of  the  given 
square. 


A                               S  B 

H K 

To  divide  AB  at  D  so  that  the  sum  of  the  squares  on  AD 
and  Z> 5  may  be  equivalent  to  the  square  on  UK. 

Construct  the  angle  ABC  equal  to  half  a  right  angle. 
With  A  as  center  and  radius  equal  to  HK,  describe  an  arc 
cutting  J5C  in  C.     Draw  CD  perpendicular  to  AB. 

The  point  D  is  the  required  point  of  division. 

To  prove  this,  draw  the  line  AC. 

The  angle  BCD,  being  equal  to  the  diiference  of  the  angles 
ADC  and  DBCy  is  equal  to  Jjalf  a  right  angle,  and  is  therefore 
equal  to  the  angle  DBC.     Hence  DC  equals  DB  (I.  62). 

Therefore  the  sum  of  the  squares  on  AD  and  DB  is  equiva- 
lent to  the  sum  of  the  squares  on  AD  and  DC-,  which  is 
equivalent  to  the  square  on  AC,  that  is  to  the  square  on  the 
given  line  HK. 

Restriction.  The  given  square  must  be  less  than  the  square  on  the 
whole  line,  and  greater  than  twice  the  square  on  half  the  line,  other- 
wise there  is  no  solution. 

87.  Note.     This  problem  can  also  be  solved  by  means  of  55. 

88.  Cor.  To  divide  a  given  line  externally  so  that  the  sum 
of  the  squares  on  the  two  segments  shall  he  equivalent  to  a 
given  square,  which  is  greater  than  the  square  on  the  given 
line. 


DIVISION  OF  A  LINE 


16a 


Medial  section, 

89.  Problem  13.  To  divide  a  given  line  into  two 
parts  such  that  the  rectangle  of  the  whole  line  and 
one  part  may  be  equivalent  to  the  square  on  the  other 
part. 

Let  AB  hQ  the  given  line,  on  which  it  is  required  to  find 
a  point  P  such  that  the  rectangle  oi  AB  and  BP  may  be 
equivalent  to  the  square  on  AP. 


D 

N 

c 

r 

i 

1 

4.^ 
1    " 
i 

.1 

! 

1 
1 

: 
.     i 

"S-. 

1    ^v. 

j 

1 

1 

i'' 

jB 

On  AB  construct  the  square  ABCD.  Bisect  AD  at  E,  and 
join  EB.  Prolong  EA,  and  take  EF  equal  to  EB.  On  AF 
construct  the  square  AFKP. 

Then  P  is  the  required  point  of  division. 

To  prove  this,  complete  the  rectangle  FEND. 

Since  AD  is  bisected  at  E  and  prolonged  to  F,  therefore 
the  rectangle  of  DF  and  AF  together  with  the  square  on  EA 
is  equivalent  to  the  square  on  EF  (52). 

Therefore  the  rectangle  DK\^  equivalent  to  the  difference 
of  the  squares  on  EF  and  EA ;  that  is,  to  the  difference  of 
the  squares  on  EB  and  EA  ;  that  is,  to  the  square  on  AB. 
Hence  the  rectangle  DK  is  equivalent  to  the  square  ABCD. 

Take  away  the  common  part  AN\  then  the  square  AKis 
equivalent  to  the  rectangle  PC. 

Now  PC  is  the  rectangle  oi  BC  and  PB,  that  is,  of  AB  and 
PB  ;  and  AK  is  the  square  on  AP. 


164  PLANE  GEOMETRY — BOOK  II 

Therefore  the  rectangle  of  AB  and  PB  is  equivalent  to  the 
square  on  AF. 

Note.  When  a  line  is  divided  as  in  this  problem,  it  is  said  to  be 
divided  in  medial  section^  for  reasons  that  appear  later  (V.  97).  The 
ancients  called  this  mode  of  division  sectio  aurea,  the  golden  section, 
on  account  of  its  important  applications. 

LOCUS   PROBLEMS 

90.  Problem  14.  To  find  the  locus  of  the  vertex  of 
a  triangle  whose  base  is  given  in  magnitude  and 
position,  and  whose  equivalent  square  is  given. 

Outline.  On  the  given  base  construct  a  rectangle  equivalent  to 
double  the  given  sqvflp|[78  (a)).  The  side  parallel  to  the  given  line, 
extended  indefinitely,  is  the  required  locus.     Prove  by  29,  33,  31. 

Ex.  On  a  given  line  construct  an  isosceles  triangle  equivalent  to  a 
given  square.     [Intersection  of  loci,  I.  253  (ex.)]. 

91.  Problem  15.  To  find  the  locus  of  the  vertex  of 
a  triangle  whose  base  is  given  in  magnitude  and 
position,  and  the  difference  of  the  squares  on  whose 
sides  is  equivalent  to  a  given  square. 

Outline  analysis.  Take  any  point  P  satisfying  the  given 
conditions,  and  from  it  draw  a  perpendicular  P  Q  to  the  base 
AB.  Show  that  any  point  on  PQ  satisfies  the  conditions 
(see  69);  and  that  any  point  not  on  PQ  does  not  satisfy 
them. 

Show  that  the  locus  is  constructed  by  dividing  AB  inter- 
nally or  externally  at  Q  so  that  the  difference  of  the  squares 
on  the  segments  shall  be  equivalent  to  the  given  square 
(internally  as  in  80  if  the  given  square  is  less  than  the 
square  on  the  given  line,  externally  as  in  82  if  greater),  and 
then  drawing  QP  perpendicular. 

Ex,  Show  how  to  construct  a  triangle,  being  given  the  base,  the 
equivalent  square,  and  the  difference  of  squares  on  sides.  (Intersection 
of  loci.) 


MAXIMA  AND  MINIMA  165 

MAXIMA  AND  MINIMA 

92.  Definition.  Of  all  the  magnitudes  of  a  certain  class 
that  fulfill  prescribed  conditions  the  greatest  is  called  the 
maximum,  and  the  least  the  minimum. 

Among  the  magnitudes  that  fulfill  the  prescribed  con- 
ditions there  may  be  a  number  of  equal  magnitudes  that 
are  each  greater  (or  less)  than  any  of  the  others ;  in  such 
case  each  of  the  equal  magnitudes  will  be  called  a  max- 
imum (or  minimum). 

93.  In  each  of  the  following  theorems  it  is  important  to 
distinguish  clearly  between  the  prescribed  conditions  that 
define  the  class  of  figures  considered,  and  the  additional 
condition  that  characterizes  a  maximum  (or  minimum)  figure 
of  the  class.  The  theorems  concerning  maxima  come  under 
one  of  the  following  type-forms,  which  are  converse  to  each 
other : 

1.  Among  all  the  magnitudes  that  fulfill  the  set  of  con- 
ditions A,  any  one  that  is  a  maximum  fulfills  the  additional 
condition  B. 

2.  Among  all  the  magnitudes  that  fulfill  the  set  of  con- 
ditions A,  any  one  that  fulfills  the  additional  condition  B  is 
a  maximum. 

The  first  form  asserts  that  the  additional  condition  B  is 
necessary  for  a  maximum  ;  the  second  form  asserts  that  the 
additional  condition  B  is  sufficient  for  a  maximum.  As  we 
shall  be  concerned  with  the  complete  conditions  for  a  max- 
imum, both  of  the  converse  theorems  will  be  considered; 
which  of  them  is  to  be  proved  first  will  depend  on  the 
nature  of  the  prescribed  conditions.  Similar  type-forms 
apply  also  to  theorems  concerning  minima. 

Greatest  triangle  having  tivo  given  sides. 

94.  Theorem  26.  //  two  sides  of  a  triangle  are 
given,  the  triangle  is  a  maximmn  ivJwn  tJie  given 
sides  include  a  right  angle. 


166 


PLANE  GEOMETRY — BOOK  II 


Let  the  triangles  ABC  and  A'bc  have 
the  sides  AB  and  BC  equal  to  the  sides 
A'b  and  BC  respectively.  Let  the  angle 
ABC  be  a  right  angle,  and  a'bc  an 
oblique  angle. 

To  prove  the  triangle  ABC  greater 
than  A'BC. 

[Prove  the  altitude  AB  greater  than  the  alti- 
tude A'D.^ 

95.  Cor.  I.  Conversely,  if  two  sides  are  given,  and  if  the  tri- 
angle is  a  maximum,  then  the  given  sides  include  a  right  angle. 

For  if  not,  a  greater  triangle  could  be  constructed  by 
making  the  included  angle  a  right  angle,  contrapy  to  the 
hypothesis  that  the  given  triangle  is  a  maximum. 

96.  Cor.  2.  Of  all  jyarallelograms  having  given  sides,  one 
that  is  rectangular  is  a  maximum;  and  conversely. 

Least  perimeter  in  equivalent  triangles. 

97.  Theorem  27.  Of  all  equivalent  triangles  having 
the  same  base,  tJiat  which  is  isosceles  ha^  the  least 
perimeter. 

Let  ABC  and  A'BC  be  equivalent  triangles  having  the 
same    base   BC,   the    first    triangle 
being  isosceles  and  the  second  not. 

To  prove  that  the  perimeter  of 
ABC  is  less  than  the  perimeter  of 
A'BC. 

Draw  CE  perpendicular  to  A  A', 
and  prolong  it  to  meet  the  prolonga- 
tion of  BA  in  D.     Join  A'n. 

Outline  proof.  Prove  in  succession:  AA'  parallel  to  BC  (33); 
angle  EAD  equal  to  EAC ;  AD  equal  to  AC;  A'D  equal  to  ^'C; 
sum  of  BA  and  AC  equal  to  BD ;  which  is  less  than  sum  of  BA'  and 
A'D ;  which  equals  the  sum  of  BA'  and  A'C.     Draw  conclusion. 


D 


MAXIMA  AND  MINIMA  167 

98.  Cor.  I.  Conversely,  of  all  equivalent  triangles  having 
the  saine  base,  that  which  has  the  least  perimeter  is  isosceles. 

[Prove  by  reductio  ad  absurdum,  as  in  95.] 

99.  Cor.  2.  Of  all  equivalent  triangles,  one  that  has  a 
minimum  perimeter  is  equilateral. 

Let  ABC  be  a  triangle  of  minimum  perimeter  belonging  to  the 
given  set  of  equivalent  triangles.  Take  BC  as  base.  Then  the  sides 
AC  and  AB  are  equal  by  98.  Similarly,  the  other  pairs  of  sides  are 
equal.     Hence  ABC  is  equilateral. 

100.  Cor.  3.  Conversely,  of  all  equivalent  triangles,  one 
that  is  equilateral  has  a  minimum  perimeter. 

Outline.  Let  ABC  be  an  equilateral  triangle  belonging  to  the  given 
set  of  equivalent  triangles.  Let  A'B'C  be  a  triangle  of  minimum 
perimeter  in  the  set.  Then  A'B'C  is  equilateral  (99).  Since  these 
two  equilateral  triangles  are  equivalent,  hence  their  sides  are  equal 
(32,  ex.).  Therefore  ABC  is  isoperimetric  with  A'B'C,  and  has  thus 
the  minimum  perimeter  in  the  set  of  equivalent  triangles. 

Greatest  surface  in  isoperimetric  triangles, 

101.  Theorem  28.  Of  all  triangles  having  a  given 
perimeter  and  a  given  base,  one  thai  is  isoscelss  has  a 
majciTnum  surface.  a 

Let  ABC  be  an  isosceles  triangle, 
and  let  ^'^C  be  any  other  triangle 
having  an  equal  perimeter  and  the  / 

same  base  BC.  // 

To    prove    the    triangle    ABC  1/ 

greater  than  ^'5 C.     Draw  ^Z)  per-        I'    / 

pendicular  to  BC,  and  A^E  parallel     f/^ 

to  5C;  join  BE,  CE.  B  D  C 

Outline  proof.  Prove  in  succession :  triangle  BEC  isosceles ; 
perimeter  of  BEC  less  than  that  of  BA'C  (97) ;  which  is  equal  to  that 
of  BAC  (hyp.)  ;  hence  BE  less  than  BA  ;  ED  less  than  AD;  triangle 
BAC  greater  than  BEC;  hence  greater  than  BA'C.  Draw  general 
conclusion. 

MCM.  ELEM.  GEOM.  —  12 


168  PLANE  GEOMETRY — BOOK  11 

102.  Cor.  I.  Conversely,  of  all  triangles  having  a  given 
perimeter  and  a  given  base,  one  that  has  a  maximum  surface 
is  isosceles. 

103.  Cor.  2.  Of  all  triangles  having  a  given  perimeter,  one 
that  has  a  maximum  surface  is  equilateral.     [Compare  102.] 

104.  Cor.  3.  Conversely,  of  all  triangles  having  a  given 
perimeter,  one  that  is  equilateral  has  a  maximum  surface. 

Rectangle  of  parts  greatest, 

105.  Theorem  29.  If  a  line  is  divided  into  any  two 
parts,  the  rectangle  of  the  parts  is  a  maximum  wJien 
the  two  parts  are  equal. 

[Divide  the  line  equally  and  unequally,  and  show  by  49  that  the 
rectangle  of  the  unequal  parts  is  less  than  that  of  the  equal  parts.  ] 

106.  Cor.  If  a  square  and  a  rectangle  have  equal  perimeters, 
the  square  is  greater  than  the  rectangle. 

Sum  of  squares  least, 

107.  Theorem  30.  //  a  line  is  divided  into  any  two 
parts,  the  sum  of  the  squares  on  tJie  parts  is  a  mini- 
mum  wlxen  the  two  parts  are  equal. 

[Show  that  the  sum  of  the  squares  on  the  unequal  parts  is  greater 
than  the  sum  of  the  squares  on  the  equal  parts  (54).] 

EXERCISES 

1.  Divide  a  given  line  into  two  parts  such  that  the  square  on  one 
of  them  may  be  double  the  square  on  the  other. 

2.  If  A,B,C^  and  D  are  four  points  in  order  on  a  line,  then  the 
rectangle  of  AD  and  BC  is  equivalent  to  the  sum  of  the  rectangles  of 
AB  and  CD,  and  of  AD  and  BC. 

3.  Divide  a  given  line  into  three  parts  so  that  the  sura  of  the  squares 
on  them  may  be  a  minimum. 

4.  Construct  a  rectangle  equivalent  to  a  given  square,  and  having 
(1)  the  sum,  (2)  the  difference  of  adjacent  sides  equal  to  a  given  line. 

5.  Show  that  42  may  be  regarded  as  a  limiting  case  of  62  ;  and  46 
of  63.     Show  that  54  and  57  may  be  regarded  as  limiting  cases  of  67. 


BOOK   III.— THE   CIRCLE 

FUNDAMENTAL  PROPERTIES 

1.  The  circle  and  its  center  and  radius  were  defined  in 
Introduction  28;  and  the  postulate  relating  to  drawing  a 
circle  was  stated  in  the  next  article. 

Since  the  circle  is  a  closed  curve,  a  line  drawn  from  any 
point  within  the  circle  to  any  point  without  intersects  the 
circle  at  least  once;  and  an  indefinite  line  drawn  through 
any  point  within  the  circle  intersects  it  at  least  twice. 

That  no  straight  line  intersects  a  circle  more  than  twice 
will  be  shown  in  theorem  4,  and  corollary. 

Pkopositions  relating  to  the  Center 

2.  Theorem  1.     A  circle  has  only  one  center. 
Let  ^5(7  be  a  circle  described  with  center  o  (post.  3). 

To  prove  that  there  is  no  other  cen- 
ter than  0. 

Suppose,  if  possible,  that  the  circle 
has  another  center  F.  a\ -^-p- 

Draw  OP,  and  let  it  meet  the  circle 
in  the  points  A  and  B. 

Since   0  is  a  center,  therefore  OA 
equals  OB.     Hence  0  is  the  mid-point  oi  AB. 

Similarly,  P  is  the  mid-point  of  AB,  which  is  impossible 
unless  0  and  P  coincide. 

Therefore  a  circle  has  only  one  center. 


170  PLANE  GEOMETRY  —  BOOK  III 

3.  Theorem  2.  If  any  two  points  are  taken  on  a 
circle,  the  perpendicular  bisector  of  the  line  joining 
them  passes  through  the  center. 

For  since  the  two  lines  drawn  from  the  center  to  the  two 
given  points  are  equal,  therefore  the  center  lies  on  the 
perpendicular  bisector  of  the  line  joining  the  two  given 
points  (I.  253). 

Ex.  What  is  the  locus  of  the  centers  of  all  the  circles  that  pass 
through  two  given  points  ? 

4.  Problem  1.     To  find  tlie  center  of  a  given  circle. 

Let  ABC  be  the  given  circle  of  which  it  is  required  to 
find  the  center. 

V 


Take  four  points  A,  B,  C,  and  D  on  the  circle,  and  let  them 
be  so  situated  that  the  lines  AB  and  CD  are  not  parallel. 
Through  the  mid-points  of  AB  and  CD  draw  the  perpendicu- 
lars EO  and  FO ;  these  lines  are  not  parallel,  for  otherwise 
the  lines  AB  and  CD  would  be  parallel.  Let  the  perpendicu- 
lars intersect  in  O. 

The  point  0  is  the  required  center. 

For  the  center  lies  on  each  of  the  lines  EOf  FO  (3). 

Therefore  the  center  is  at  0,  the  intersection  of  the  two 
perpendiculars. 

Note.  This  construction  may  be  regarded  as  an  application  of  the 
method  of  intersection  of  loci  (I.  256). 

Ex.  Being  given  any  portion  of  a  circle,  show  how  to  find  the 
center,  and  how  to  complete  the  circle. 


FUNDAMENTAL  PROPERTIES  171 

5.  Theorem  3.  The  line  joining  any  point  to  the 
center  of  a  circle  is  less  than,  equal  to,  or  greater 
than  the  radius,  according  as  the  point  is  within, 
on,  or  without,  the  circle. 

Let  ABC  be  a  circle  whose  center  is  0.  Let  P,  Q,  R  be 
any  points  within,  on,  and  without,  the  circle,  respectively. 

To  prove  that  OP  is  less  than  a         ^ ^x,,^^^ 

radius,  OQ  equal  to  a  radius,  and  OR      /^  >.        j^ 

greater  than  a  radius.  '  /  ^ — ^ 

Extend  OP  to  meet  the  circle  in  ^4,    I  tX.         I 

and  let  OR  cut  the  circle  in  c.  \  p«      ^/q 

The   lines    OA,    OQ,   and   OC    are        ^-^_L^''^^ 
equal,  being  radii. 

But  OP  is  less  than  OA ;  and  07?  is  greater  than  OC. 

Therefore  OP,  OQ,  OR  are  respectively  less  than,  equal  to, 
and  greater  than,  the  radius. 

6.  Cor.  I.  A  point  is  luithin,  on,  or  without,  the  circle, 
according  as  the  line  joining  it  with  the  center  is  less  than, 
equal  to,  or  greater  than,  the  radius. 

7.  Cor.  2.  The  locus  of  a  point,  such  that  its  join  to  a  given 
point  shall  be  equal  to  a  given  line,  is  a  circle  described  with 
the  given  point  as  center  and  the  given  line  as  radius. 

Intersections  of  Line  and  Circle 

8.  Theorem  4.  //  any  two  points  of  a  circle  are 
joined  by  a  straight  line,  all  points  of  this  line 
situated  between  the  given  points  lie  within  the 
circle;  and^  all  points  in  the  line  extended  either 
way  beyond  the  given  points,  lie  without  the  circle. 

Let  A,  BhQ  any  two  points  on  the  circle ;  D  a  point  in  the 
line  AB,  situated  between  A  and  B ;  F  any  point  on  the  line 
AB  extended  either  way. 


172 


PLANE  GEOMETRY  —  BOOK  III 


To  prove  that  D  is  within,  and  F  without,  the  circle. 

Find  O,  the  center  of  the 
circle  (4).  Join  OAy  OBy  OD, 
0F\  and  extend  OD  through 
D  to  meet  the  circle  in  E. 

Since  the  triangle  OAB  is 
isosceles,  OD  is  less  than  05, 
and  OF  is  greater  than  OB 
(1.86). 

Then  since  OD  is  less  than  the  radius,  the  point  D  is 
within  the  circle ;  and  since  OF  is  greater  than  the  radius, 
the  point  F  is  without  the  circle  (5). 


9.   Cor.     A  straight  line  cannot  meet  a  circle  in  more  than 
two  points. 

Chorda  and  secants, 

10.  Definitions.  A  straight  line  that  meets  a  circle  in  two 
points  is  called  a  secant.  The  portion  of  a  secant  included 
within  the  circle  is  called  a  chord.  A  chord  that  passes 
through  the  center  is  called  a  diameter. 


11.  Theorem  5.  The  perpendicular  from  the  center 
to  a  chord  bisects  it;  and,  conversely,  tJie  line  drawn 
from  the  center  to  the  mid-point  of  a  chord  is  per- 
pendicular to  the  chord. 

Let  ABhQ  any  chord  of  the  circle  ABC, 
whose  center  is  0.  Let  OD  be  the  per- 
pendicular from  0  to  AB. 

To  prove  that  AB  is  bisected  at  D. 
(Use  I.  99.) 

Converse.     Let  uiB  be  bisected  at  D. 

To  prove  OD  perpendicular  to  AB.     (Use  I.  66.) 


FUNDAMENTAL  PROPERTIES  173 

Ex.  1.  The  line  which  bisects  perpendicularly  one  of  two  parallel 
chords  of  a  circle  also  bisects  the  other  perpendicularly. 

Ex.  2.  If  a  line  intersects  two  concentric  circles,  the  intercepts 
between  the  circles  are  equal. 

Ex.  3.   Any  diameter  of  a  circle  is  an  axis  of  symmetry. 

12.  Theorem  6.  The  least  line  that  can  he  drawn 
from  a  given  point  to  a  given  circle  is  a  segment  of  the 
line  that  passes  through  the  center,  and  the  ejotremities 
of  this  segment  are  at  the  same  side  of  the  center. 

Let  ABCD  be  a  circle  whose  center  is  0.  Let  P  be  any 
point,  either  without  the  circle  (fig.  1),  or  within  the  circle 
(fig.  2).  Let  the 
line  PO  meet  the 
circle  in  the  points 
A  and  D,  of  which  A 
is  at  the  same  side 
of  the  center  as  P  is. 
Let  PBC  be  any 
other  secant  through 

,.,,.,  Fig.  1  Fig.  2 

P  meeting  the  circle 

in  the  points  B  and  G,  of  which  B  is  at  the  same  side  of 

the  mid-point  of  the  chord  i?C'  as  P  is. 

To  prove  that  PA  is  less  than  PB. 

In  the  triangle  POB  (fig.  1),  the  side  OP  is  less  than  the 
sum  of  the  sides  OB  and  PB  (I.  87). 

Taking  away  the  equals  OA  and  OB,  it  follows  that  PA  is 
less  than  PB. 

Again,  in  the  triangle  POB  (fig.  2),  the  sum  of  OP  and 
PB  is  greater  than  OB,  and  therefore  greater  than  OA. 

Taking  away  the  common  part  OP,  it  follows  that  PB  is 
greater  than  PA,  that  is,  PA  is  less  than  PB. 

13.  Cor.  I.  The  greatest  line  from  any  point  to  the  circle 
is  a  ^portion  of  the  same  secant  as  the  least  line  is. 

14.  Cor.  2.     The  greatest  chord  of  a  circle  is  a  diameter. 


174  PLANE  GEOMETRY — BOOK  III 

PROPERTIES  OF  EQUAL  CIRCLES 
Conditions  fob  Coincidence 

15.   Theorem  7.     Circles  of  equal  radii  are  equal. 

Let  ABC  J  a'b'c'  be  two  circles   of  equal   radii,  whose 
centers  are  0  and  O'. 

c  & 


To  prove  that  the  circles  are  equal. 

Let  P  be  any  point  on  the  circle  ABC. 

Place  the  circle  ABC  upon  a'b'c'  so  that  the  center  O 
may  coincide  with  the  center  O'.  Let  the  point  P  fall 
on  P'. 

The  line  O'p'  equals  OP,  being  coincident  with  it;  and 
OP  equals  the  radius  of  the  circle  A'b'c'  (hyp.). 

Therefore  O'P'  equals  the  radius  of  the  circle  a'b'c'. 

Hence  P'  is  on  the  circle  A'b'c'  (6). 

Thus  any  point  of  the  circle  ABC  will  fall  on  the  circle 
A'B'C'. 

In  like  manner,  any  point  of  the  circle  A  'b'c'  will  coincide 
with  a  point  of  the  circle  ABC. 

Hence  the  two  circles  coincide  in  all  their  parts,  and  are 
therefore  equal. 

16.  Cor.  I.     If  tivo  circles  are  equal,  their  radii  are  equal. 
What  are  the  contraposites  of  15  and  16  ? 

17.  Cor.  2.  Two  circles  which  have  one  point  in  common, 
and  which  have  the  same  center,  coincide  throughout. 


PROPERTIES   OF  EQUAL   CIRCLES  175 

18.  Cor.  3.  Two  circles  which  have  07ie  point  in  common, 
and  which  do  not  coincide  throughout,  have  not  the  same  center. 

How  is  this  corollary  related  to  the  preceding  ? 

19.  Cor.  4.  Two  circles  which  have  the  same  center  and 
have  unequal  radii  have  no  point  in  common. 

20.  Theorem  8.  Through  three  given  points  not  in 
the  same  straight  line,  one,  and  only  one,  circle  can 
-pass. 

Let  A,  B,  C  be  three  given  points  not  in  the  same  straight 
line. 

First,  to  prove  that  a  circle  can  pass 
through  A,  B,  C. 

Find  the  point  0  such  that  the  three 
lines  OA,  OB,  OC  are  equal  (I.  257). 

The  circle  described  with  0  as  center 
and  a  radius  equal  to  OA  passes  through 
A,  B,  C,  and  thus  fulfills  the  requirements. 

]N"ext,  to  prove  that  only  one  circle  can  pass  through 
A,  B,  C. 

The  point  O  is  the  only  point  such  that  OA,  OB,  OC  are 
equal  (I.  257). 

Therefore  all  the  circles  passing  through  A,  B,  G  have  the 
same  center  and  equal  radii ;  and  hence  they  coincide  (17). 

Therefore  there  is  only  one  circle  passing  through  A,  B,  G. 

21.  Cor.  I.  Two  circles  having  three  poiyits  in  common 
coincide. 

Show  that  this  is  only  another  statement  of  the  second  part  of  20. 

22.  Cor.  2.  Two  circles  that  do  not  coincide  do  not  meet  in 
more  than  tivo  points. 

How  is  this  corollary  related  to  the  preceding  ? 


176  PLANE  GEOMETRY — BOOK  III 

23.  Cor.  3.  If  from  any  point  within  a  circle  more  than 
two  lines  drawn  to  the  circle  are  equal,  then  that  point  is  the 
center. 

Outline.  Show  that  the  opposite  of  this  conclusion  would  lead  to 
the  opposite  of  I.  257. 

24.  Cor.  4.  If  two  triangles  be  equal,  the  circle  passing 
through  the  vertices  of  one  is  equal  to  the  circle  parsing  through 
the  vertices  of  the  other. 

25.  Cor.  5.  Two  circles  cannot  have  a  common  portion 
without  coinciding  throughout. 

Arcs  and  Central  Angles 

26.  Definitions.  An  arc  is  part  of  a  circle.  Two  arcs 
which  together  make  up  the  whole  circle  are  said  to  be 
conjunct  arcs.  When  two  conjunct  arcs  are  equal,  each  is 
called  a  semicircle.  When  two  conjunct  arcs  are  unequal, 
the  greater  is  called  the  major  conjunct  arc  and  the  less 
the  minor  conjunct  arc. 

An  angle  at  the  center  of  a  circle  formed  by  two  radii  is 
called  a  central  angle.  Of  the  two  conjunct  central  angles 
formed  by  the  same  two  radii,  the  less  is  called  the  jninor 
and  the  other  the  major. 

The  two  conjunct  central  angles  are  said  to  stand  on 
the  two  arcs  intercepted  by  the  sides,  the  minor  angle  on 
the  minor  arc,  and  the  major  angle  on  the  major  arc. 
In  the  same  case  the  minor  arc  is  said  to  subtend  the 
minor  angle,  and  the  major  arc  the  major  angle.  An  arc 
subtending  a  central  right  angle  is  called  a  quadrant. 

A  sector  is  a  figure  composed  of  an  arc  and  the  two  radii 
drawn  to  its  extremities.  The  central  angle  formed  by  the 
radii  is  called  the  angle  of  the  sector ;  and  the  sector  is 
said  to  contain  the  central  angle. 

In  any  two  circles,-  arcs  that  subtend  equal  central  angles 
are  said  to  be  similar,  and  so  are  the  corresponding  sectors. 


PROPERTIES   OF  EQUAL   CIRCLES  177 

27.  Theorem  9.  In  equal  circles,  or  in  the  same 
circle,  equal  central  angles  stand  on  equal  arcs. 

Let  ABC,  a'b'c'  be  equal  circles  whose  centers  are  0,  O'. 
Let  A  OB,  a'o'b'  be  equal  central  angles. 

To  prove  that  the 
arc  AB  is  equal  to  the 
arc  A  'b'. 

Let  the  circle  ABC 
be  applied  to  A'b'c' 
so  that  the  radius  OA 
coincides  with  the 
equal  radius  O'a',  and 
the  angle  AOB  with  its  equal  angle  a'o'b'. 

Then  the  radius  OB  coincides  with  the  equal  radius  O'B' -, 
and  the  circle  ABC  with  the  circle  A'b'c'  (15). 

Hence  the  arc  AB  coincides  with  the  arc  A'b';  and  these 
arcs  are  therefore  equal. 

Ex.  1.  A  diameter  divides  a  circle  into  two  equal  arcs ;  and  two 
diameters  at  right  angles  divide  it  into  four  equal  arcs. 

Ex.  2.  All  quadrants  of  the  same  circle  are  equal ;  and  each  is 
equal  to  one  fourth  of  the  circle. 

27  (fl).  Cor.  In  equal  circles,  or  in  the  same  circle,  sectors 
which  include  equal  central  angles  are  equal. 

Note.  As  it  is  evident  that  the  theorems  proved  for  equal  circles 
are  also  true  for  the  same  circle,  the  words  "  or  in  the  same  circle" 
will  usually  be  omitted. 

28.  Definition.  As  equal  circles  have  equal  radii,  any  two 
arcs  of  equal  circles  will  be  called  equira^dial  arcs;  and 
any  two  sectors  of  equal  circles  will  be  called  equiradial 
sectors. 

29.  Comparison  of  equiradial  arcs.  Two  equiradial  arcs  are 
compared  in  the  same  way  as  two  line-segments  are  com- 
pared, viz.  by  transferring  one  so  that  an  extremity  falls  on 
an  extremity  of  the  other  and  so  that  one  of  the  arcs  may 


178  PLANE  GEOMETRY  —  ROOK  III 

coincide  with  the  whole  or  part  of  the  other.  The  terms 
equal,  greater,  and  less  are  then  applied  in  accordance 
with  the  general  definitions  (Introd.  35). 

The  surfaces  of  two  equiradial  sectors  are  compared  in  the 
same  way  as  two  angles  are  (I.  11). 

30.  Theorem  10.  In  equal  circles  two  unequal 
central  angles  stand  on  unequal  arcs,  tlie  greater 
angle  standing  on  the  greater  arc. 

In  the  equal  circles  ABC^  A^B'c\  let  the  central  angle  AOB 
be  less  than  the  central  angle  A'o'c'. 

C 


To  prove  that  the  arc  AB  is  less  than  the  arc  A'c'. 

Draw  the  line  O'b'  cutting  off  from  the  greater  angle  A'o'c' 
a  part  A' o'b'  equal  to  the  less  angle  AOB  (I.  77). 

Then  the  arc  a'b'  equals  the  arc  AB  (27). 

Therefore  the  arc  A  'c',  being  greater  than  its  part  A  'b',  is 
greater  than  the  arc  AB. 

Combined  statement, 

31.  Cor.  I.  In  two  equal  circles j  according  as  a  central 
angle  of  one  is  greater  thauy  equal  to,  or  less  than,  a  central 
angle  of  the  other,  so  is  the  arc  subtending  the  first  greater  than, 
equal  to,  or  less  than,  the  arc  subtending  the  second. 

32.  Cor.  2.  In  two  equal  circles,  according  as  a  central 
angle  of  one  is  greater  than,  equal  to,  or  less  than,  a  central 
angle  of  the  other,  so  is  the  sector  containing  the  first  greater 
than,  equal  to,  or  less  than,  the  sector  containing  the  second. 


PROPERTIES   OF  EQUAL   CIRCLES  179 

33.  Cor.  3.  According  as  a  central  angle  is  greater  than, 
equal  to,  or  less  than,  a  right  angle,  so  is  the  opposite  arc  greater 
than,  equal  to,  or  less  than,  a  quadrant  ;  and  conversely. 

34.  Cor.  4.  According  as  a  central  angle  is  greater  than, 
equal  to,  or  less  thayi,  a  straight  angle,  so  is  the  opposite  arc 
greater  than,  equal  to,  or  less  than,  a  semicircle  ;  and  conversely. 

Converse  of  31, 

35.  Theorem  11.  In  two  equal  circles,  according  as 
an  arc  of  one  is  greater  than,  equal  to,  or  less  than, 
an  arc  of  the  other,  so  is  the  central  angle  standing 
on  the  first  greater  than,  equal  to,  or  less  than,  the 
central  angle  standing  on  the  second. 

Addition  of  equiradial  arcs,  or  sectors, 

36.  Definitions.  The  sum  of  two  equiradial  arcs  is  the 
arc  obtained  by  laying  off  on  any  equiradial  circle  two  arcs 
respectively  equal  to  the  given  arcs,  so  as  to  have  a  common 
extremity  without  overlapping.  This  sum  is  called  the 
result  of  adding  the  two  arcs. 

The  supi  of  three  or  more  equiradial  arcs  is  the  result  of 
adding  the  third  arc  to  the  sum  of  the  first  two,  and  so  on. 

The  arc  obtained  by  adding  two  equal  arcs  is  called  the 
double  of  either,  and  each  of  the  latter  is  said  to  be  half  of 
the  former. 

The  difference  of  two  unequal  equiradial  a^cs  is  the  arc 
which  must  be  added  to  the  less  to  produce  the  greater. 

The  sum  of  two  or  more  equiradial  sectors  is  the  whole 
sector  obtained  by  placing  them  so  that  their  angles  are 
adjacent  in  succession  (I.  9).  The  words  double,  half,  dif- 
ference, are  applied  to  equiradial  sectors  in  the  usual  way. 

37.  Magnitudes  directly  comparable.  We  have  met  with 
four  kinds  of  magnitudes  such  that  two  of  the  same  kind 
may  always  be  directly  compared  by  superposition  (without 
previous  dissection).  They  are:  line-segments  (I.  2); 
angles  (I.  9) ;  equiradial  arcs  (29) ;  equiradial  sectors. 


180  PLANE  GEOMETRY  —  BOOK  III 

38.  Axioms  of  equality  and  inequality.  The  axioms  given 
in  I.  24-32  for  line-segmeuts  and  angles  may  now  be  proved 
for  equiradial  arcs  and  sectors  as  direct  inferences  from  the 
definitions  given  in  29,  36. 

39.  Additional  principles  of  equality.  The  following  addi- 
tional principles  can  be  easily  proved  for  these  four  kinds  of 
magnitudes  and  will  be  of  frequent  use : 

a.  The  half  of  the  sum  of  two  magnitudes  is  equal  to  the 
sum  of  their  halves. 

b.  The  double  of  the  difference  of  two  magnitudes  is  equal 
to  the  difference  of  their  doubles. 

c.  The  half  of  the  difference  of  two  magnitudes  is  equal 
to  the  difference  of  their  halves. 

These  are  left  as  exercises.  The  corresponding  proofs  for  "equiva- 
lence "  in  II.  13-17  may  be  consulted. 

40.  Theorem  12.  The  sum  of  two  or  more  arcs  of 
equal  circles  subtends  a  central  angle  equal  to  the 
sum  of  the  central  angles  standing  on  the  separate 
arcs.    [Use  35.] 

40  (a).  Cor.  i.  The  difference  of  two  arcs  of  equal  circles 
subtends  a  central  angle  equal  to  the  difference  of  the  central 
angles  standing  on  the  separate  arcs. 

p]x.  1.  In  equal  circles,  if  one  central  angle  is  three  times  another, 
the  arc  opposite* the  first  is  three  times  the  arc  opposite  the  second. 

Ex.  2.     Show  how  to  bisect  a  given  arc,  or  a  given  sector. 

Ex.  3.  Show  how  to  trisect  a  given  quadrant,  or  semicircle,  or 
circle  (I.  129,  exs.). 

41.  Definition.  Two  equiradial  arcs  are  said  to  be  cottv- 
pleniental,  supplemental,  or  conjunct,  according  as 
their  sum  is  equal  to  a  quadrant,  a  semicircle,  or  a  circle. 

41  (a).  Cor.  2.  In  equal  circles,  according  as  two  central 
angles  are  comj)lemental,  supplemental,  or  conjunct,  so  are  the 
opposite  arcs  complemental,  supplemental ^  or  conjunct. 


PROPERTIES   OF  EQUAL   CIRCLES 


181 


Arcs  and  Chords 

42.  Definitions.  The  line  joining  the  extremities  of  an 
arc  is  called  the  chord  of  the  arc,  and  the  chord  is  said 
to  subtend  the  arc.  Every  chord  subtends  two  arcs,  one 
on  each  side  of  it.  If  the  chord  is  not  a  diameter,  the  two 
subtended  arcs  are  unequal,  and  the  greater  is  called  the 
major  arc  subtended  by  the  chord,  and  the  less  the  minor. 
When  the  "  arc  subtended  by  a  chord  "  is  mentioned  without 
qualification,  the  minor  arc  will  be  understood,  that  is,  the 
one  less  than  a  semicircle. 

43.  Theorem  13.  In  equal  circles,  equal  arcs  have 
equal  chords. 

Let  the  equal  circles  ABC,  a'b'c'  have  the  arcs  AB,  and 
A'b'  equal. 

To  prove  that  the 
chords  AB  and  A'b' 
are  equal. 

Let  0,  o'  be  the 
centers  of  the  circles. 
Join  O J,  OB,0'A',  O'B'. 

Since  the  arcs  AB, 
a'b'  are  equal,  therefore,  whether  these  are  major  or  minor 
arcs,  the  angles  AOB,  A'o'b'  of  the  triangles  AOB,  A' O'B' 
are  equal  (35). 

Hence  these  triangles,  having  two  sides  and  the  included 
angle  in  each  respectively  equal,  have  their  bases  AB  and 
A'b'  also  equal. 

44.  Theorem  14.  In  equal  circles,  of  two  unequal 
minor  arcs  the  greater  has  the  greater  chord ;  and  of 
two  unequal  major  arcs  the  greater  has  the  less  chord. 

Let  the  equal  circles  ABC,  a'b'c'  have  the  minor  arc  AB 
greater  than  the  minor  arc  A'B'. 


182  PLANE  GEOMETRY — BOOK  III 

To  prove  that  the  chord  ^-B  is  greater  than  the  chord  a'b\ 

Let  0,  O'  be  the  cen- 
ters   of    the    circles.  C  ^^-^..^^ 
Joino^,  05, 0'J',  0'^'.         ^^^       \            /  ^ 

Since  the  minor  arc 
AB  \^  greater  than  the 
minor  arc  A^B\  the 
central  angle  AOB  is 
greater  than  the  cen- 
tral angle  ^'o'J5'  (35). 

Hence,  in  the  triangles  AOB,  A'o'b',  the  sides  OA,  OB  are 
equal  to  the  sides  o'A'y  o'b',  and  the  angle  AOB  is  greater 
than  the  angle  a'o'b'  -,  therefore  the  base  ^J5  is  greater  than 
the  base  a'b'  (I.  91). 

Next,  let  the  major  arc  ACB  be  less  than  the  major  arc 
A'C'B'. 

To  prove  that  the  chord  ^B  is  greater  than  the  chord  ^'B'. 

Subtracting  the  unequal  major  arcs  from  the  equal  circles, 
it  follows  that  the  minor  arc  AB  is  greater  than  the  minor 
arc  ^'5' (38;  1.31). 

Therefore,  by  the  preceding  case,  the  chord  AB  is  greater 
than  the  chord  A'b'. 

Combined  statement, 

44  (a).  Cor.  In  equal  circles,  according  as  a  minor  arc  of 
one  circle  is  greater  than,  equal  to,  or  less  than,  a  minor  arc 
of  the  other,  so  is  the  chord  of  the  first  arc  greater  than,  equal 
to,  or  less  than,  the  chord  of  the  other  arc.     [Combine  43,  44.] 

Converse  statement, 

45.  Theorem  15.  In  equal  circles,  according  as  one 
chord  is  greater  than,  equal  to,  or  less  than,  another 
chord,  so  is  the  minor  arc  subtended  hy  tJve  first  clwrd 
greater  than,  equal  to,  or  less  than,  the  minor  arc 
subtended  by  the  second  chord.     [Rule  of  Conversion.] 


PROPERTIES   OF  EQUAL   CIRCLES 


183 


45  (a).  Cor.  In  equal  circles,  according  as  one  chord  is 
greater  than,  equal  to,  or  less  than,  another  chord,  so  is  the 
major  arc  subtending  the  Jirst  chord  less  than,  equal  to,  or 
greater  than,  the  major  arc  subtending  the  second  chord. 

First  combine  43  and  the  second  part  of  44  into  a  triple  state- 
ment ;  and  then  apply  Rule  of  Conversion. 


Chords  and  Central  Perpendiculars 

46.  Theorem  16.  In  equal  circles,  the  perpendicu- 
lars from  the  centers  to  equal  chords  are  equal. 

In  the  equal  circles  ABC,  A'b'c',  let  OM,  O'm'  be  the  per- 
pendiculars from  the  centers  to  the  two  equal  chords  AB, 
A'B'. 

C  C 


To  prove  that  the  perpendiculars  DM  and  o'm'  are  equal. 

The  two  right  triangles  0AM,  O'a'm'  have  their  hy- 
potenuses equal,  and  the  sides  AM,  A'm'  equal,  being  halves 
of  equal  chords  (11). 

Therefore  the  third  sides  OM  and  o'm'  are  equal  (I.  98). 

47.  Theorem  17.  In  equal  circles,  if  perpendicu- 
lars are  drawn  from  the  centers  to  two  unequal 
chords,  the  perpendicular  drawn  to  the  less  chord  is 
the  greater. 

In  the  equal  circles  ABC,  A'b'c',  let  the  chord  AB  of  the 
first  be  less  than  the  chord  A'b'  of  the  second.  Let  OM, 
o'm'  be  the  perpendiculars  from  the  centers  to  these  chords. 


MCM.   ELEM.   GEOM. 


13 


184  PLANE  GEOMETRY  —  BOOK  III 

C  O' 


To  prove  that  DM  is  greater  than  O^M^. 

The  chord  AB  is  less  than  ^'-B';  and  AM,  a'm'  are  halves 
of  ^5,^  V  (11). 

Therefore  ^Af  is  less  than  a'm'  (I.  32,  ax.  14). 

Then,  in  the  right  triangles  GAM,  O'a'm'  the  hypotenuses 
are  equal,  and  the  side  ^^  of  the  first  is  less  than  the  side 
a'm'  of  the  second. 

Now  the  square  on  OM  is  equivalent  to  the  difference  of  the 
squares  on  OA  and  AM  (II.  61);  and  the  square  on  O'm'  is 
equivalent  to  the  difference  of  the  squares  on  o'A'  and  A'm'. 
Subtracting  unequals  from  equals,  it  follows  that  the  square 
on  OM  is  greater  than  the  square  on  o'm'  (I.  31,  ax.  11,  and 
II.  15  (a)).     Hence  OM  is  greater  than  o'm'  (II.  24). 

Cotnbined  statement, 

48.  Cor.  In  equal  circles,  if  perpendiculars  are  drawn  from 
the  centers  to  any  two  chords,  then  according  as  the  first  chord 
is  greater  than,  equal  to,  or  less  than,  the  second  chord,  so  is 
the  perpendicular  to  the  first  less  than,  equal  to,  or  greater  than, 
the  perpendicular  to  the  second. 

Converse  statement, 

49.  Theorem  18.  In  equal  circles,  if  perpendicu- 
lars are  draivn  from  the  centers  to  any  two  chords, 
tlien  according  as  the  first  perpendicular  is  less  than, 
equal  to,  or  greater  than,  the  second,  so  is  the  first 
chord  greater  than,  equal  to,  or  less  than,  the  second 
chord. 


PROPERTIES   OF  EQUAL   CIRCLES  185 

Concerning  Order-theorems 

50.  Order  of  size  of  a  pair.  When  two  magnitudes  A  and 
B  of  the  same  kind  are  compared  as  to  whether  A  is  greater 
than,  equal  to,  or  less  than,  B,  the  statement  of  the  result  is 
called  the  order  of  size  of  the  pair  {A,  b). 

When  the  two  magnitudes  are  named  or  written  in  any 
order,  this  order  will  be  called  ascending  when  the  first  is 
less  than  the  second,  descending  when  the  first  is  greater 
than  the  second,  and  indifferent  when  the  two  magnitudes 
are  equal. 

To  illustrate  the  convenience  of  this  phraseology,  the  two 
converse  propositions  in  I.  82,  84,  each  containing  a  triple 
statement,  may  be  enunciated  thus : 

In  a  triangle,  the  pair  of  sides  {a,  h)  and  the  pair 
of  opposite  angles  {A,  B)  are  in  the  same  order  of  size. 

This  asserts  that  if  either  pair  is  in  ascending  order,  so  is 
the  other;  if  either  pair  is  in  descending  order,  so  is  the 
other ;  and  if  the  order  of  either  pair  is  indifferent,  so  is  that 
of  the  other.  Hence  this  statement  includes  six  different 
statements. 

A  theorem  that  compares  the  order  of  size  of  two  pairs 
of  magnitudes  is  called  an  order-theorem. 

As  another  illustration,  I.  94  and  its  converse  may  be 
enunciated  as  an  order-theorem ;  thus  : 

If  two  triangles  have  two  sides  of  one  equal  to  two 
sides  of  the  other,  then  the  pair  of  included  angles 
{A,  A')  and  the  pair  of  opposite  sides  {a,  a')  are  in 
the  same  order  of  size. 

Similarly,  the  two  triple  statements  in  II.  23,  24  may  be 
abbreviated  as  follows : 

If  two  rectangles  have  equal  altitudes,  then  the 
pair  of  rectangles  (H,  It')  and  the  pair  of  bases  (b,  b') 
are  in  the  same  order  of  size. 

What  are  the  six  statements  included  in  this  enunciation  ? 


186  PLANE  GEOMETRY  —  BOOK  III 

51.  Summary  of  order-theorems  in  equal  circles.  All  the 
order-theorems  proved  for  equal  circles  may  be  conveniently 
summarized  thus : 

In  equal  circles  the  following  pairs  of  magnitudes 
are  all  in  the  same  order  of  size: 

(1)  Any  two  central  angles  {A,  A'); 

(2)  The  two  opposite  arcs  {a,  a') ; 

(3)  The  containing  sectors  {S,  S') ; 

(4)  The  subtending  chords  {c,  c') ; 

(6)   The  central  perpendiculars  ip',  p). 

EXERCISES 

1.  The  chords  cut  off  by  two  equal  circles  from  any  line  parallel  to 
the  line  joining  their  centers  are  equal. 

2.  The  chords  cut  off  by  two  equal  circles  from  any  line  drawn 
through  the  mid-point  of  the  line  joining  their  centers  are  equal. 

3.  If  a  line  is  drawn  intersecting  two  concentric  circles,  the  two 
parts  intercepted  between  the  circles  are  equal. 

4.  If  two  equal  chords  are  drawn  in  a  circle,  the  two  portions 
intercepted  by  a  concentric  circle  are  equal. 

5.  Find  the  locus  of  the  mid-points  of  equal  chords  of  a  circle. 

6.  Find  the  locus  of  the  points  of  trisection  of  equal  chords. 

7.  Through  a  given  point  inside  a  circle  draw  the  least  chord. 

ANGLES  IN  SEGMENTS 

52.  Definitions.    The  figure  formed  by  an  arc  of  a  circle 

and  its  chord  is  called  a  segment  of  the  circle. 

Two  segments  are  said  to  be  conjunct  when  their  arcs 
are  conjunct  (41).  Segments  of  different  circles  are  said 
to  be  similar  if  their  arcs  are  similar  (26). 

A  segment  not  semicircular  is  called  a  major  or  a  minor 
segment,  according  as  its  arc  is  a  major  or  a  minor  arc. 

The  angle,  not  convex,  formed  by  any  two  chords  that 
meet  on  the  circle,  is  called  an  inscribed  angle,  and  is  said 
to  stand  upon  the  arc  which  is  between  the  sides  of  the 
angle.    In  the  same  case  the  arc  is  said  to  subtend  the  angle. 


ANGLES  IN  SEGMENTS 


187 


The  angle,  not  convex,  formed  by  two  straight  lines 
drawn  from  a  point  in  the  arc  of  a  segment  to  the  extremi- 
ties of  its  chord  is  called  an  angle  in  the  segment. 


Angles  standing  on  the  Same  Arc 

53.  Theorem  19.  An  inscribed  angle  is  equal  to 
half  the  central  angle  standing  on  the  same  arc. 

Let  BCA  be  an  inscribed  angle  stand- 
ing on  the  arc  AB.  Let  BOA  be  the 
central  angle  standing  on  the  same  arc. 

To  prove  the  angle  ACB  equals  half 
the  angle  AOB. 

First  let  the  center  0  lie  on  one  of 
the  sides  of  the  angle,  say  AC. 

Since  the  radii  OB  and  OC  are  equal,  the  angles  OCB  and 
0£Care  equal  (I.  59). 

Therefore  the  angle  OCB  equals  half  the  sum  of  the 
angles  OCB  and  OBC. 

Now  the  sum  of  the  angles  OCB  and  OBC  equals  the 
exterior  angle  AOB  (I.  128). 

Therefore  the  angle  OCB  equals  half  the  angle  AOB, 

Next  let  0  lie  within  the  angle  ACB. 


Draw  AO,  BO,  CO.     Prolong  GO  to  meet  the  circle  in  Z). 

By  the  previous  case  the  angle  ACO  is  half  the  angle 
AOD,  and  OCB  is  half  DOB.  Hence  the  whole  angle  ACB  is 
half  the  whole  angle  AOB  standing  on  the  same  arc  (39,  a). 


188 


PLANE  GEOMETRY — BOOK  III 


Lastly  let  the  center  0  lie  without  the 
angle  AGB. 

Draw  AO^  BOy  CO.  Prolong  CO  to 
meet  the  circle  in  D. 

By  the  first  case  the  angle  ACD  is 
half  the  angle  ^0/>,an(i  BCD  is  half  BOD, 

Therefore  the  remaining  angle  BCA  is 
half  the  remaining  angle  BOA  (39,  c). 

Ex.     In  equal  circles,  equal  inscribed  angles  stand  on  equal  arcs. 

54.  Theorem  20.  Angles  in  the  same  segment  are 
equal. 

Let  AC  By  AC'b  be  angles  in  the  same  segment  of  the  circle 
ABc'C  whose  center  is  0. 

To  prove  that  the  angles  ACB  and 
AC'B  are  equal. 

Draw  the  lines  OA  and  OB. 

Since  the  angles  ACB  and  AC'B  stand 
on  the  same  arc  ADBj  they  are  halves 
of  the  same  central  angle  AOB  stand- 
ing on  the  arc  ADB  (53). 

Therefore  the  angles  ACB  and  AC'B  are  equal  (I.  28). 

Note.  Since  all  angles  in  the  same  segment  are  equal,  any  one  of 
them  may  be  called  the  angle  contained  in  the  segment. 

55.  Cor.  The  angle  subteiided  by  the  chord  of  a  segment  at 
a  point  within  the  segment  is  greater  than  the  angle  in  the  seg- 
ment ;  and  the  angle  subtended  at  a  point  witJiout  the  segment 
and  on  the  same  side  of  the  chord  as  the  segment,  is  less  than 
the  angle  in  the  segment.     (Use  I.  79,  and  ex.  9,  p.  44.) 

Ex.  1.  In  equal  circles,  two  segments  that  contain  equal  angles  are 
equal.     (Prove  the  arcs  equal,  and  superpose.) 

Ex.  2.  In  any  two  circles,  segments  that  contain  equal  angles  are 
similar. 


ANGLES  IN  SEGMENTS 


189 


Species  of  Inscribed  Angle 

56.  Theorem  21.  According  as  the  arc  of  a  segment 
is  greater  than,  equal  to,  or  less  than,  a  seinicircle,  so 
is  the  angle  in  the  segment  less  than,  equal  to,  or 
greater  than,  a  right  angle. 

Let  ACB  be  an  angle  in  the  segment  ABC. 


To  prove  that  ACB  is  less  than,  equal  to,  or  greater  than, 
a  right  angle,  according  as  the  arc  A  CB  is  greater  than,  equal 
to,  or  less  than,  a  semicircle. 

Take  the  center  0,  and  draw  OA,  OB. 

According  as  the  arc  ACB  is  greater  than,  equal  to,  or  less 
than,  a  semicircle,  so  is  the  conjunct  arc  ADB  less  than,  equal 
to,  or  greater  than,  a  semicircle,  and  thus  the  central  angle 
AOB  standing  on  the  arc  ABB  is  less  than,  equal  to,  or  greater 
than,  a  straight  angle  (34). 

Now  the  angle  ACB  is  half  the  central  angle  AOB  standing 
on  the  same  arc  (53). 

Therefore,  according  as  the  arc  ACB  is  greater  than,  equal 
to,  or  less  than,  a  semicircle,  the  angle  ACB  is  less  than, 
equal  to,  or  greater  than,  a  right  angle. 


57.  Theorem  22.  According  as  the  angle  in  a  seg- 
ment is  greater  than,  equal  to,  or  less  than,  a  right 
angle,  so  is  the  arc  of  the  segment  less  than,  equal  to, 
or  greater  than,  a  semicircle.     (Rule  of  Conversion.) 


190  PLANE  GEOMETRY — BOOK  III 

Angles  in  Conjunct  Segments 

58.   Theorem  23.     The  angles  in  two  conjunct  seg- 
ments are  supplemental. 

Let  AB  be  the  chord  of  the  two  conjunct  segments  ABC  and 
ABB.   Let  ACBy  ABB  be  angles  in  these  segments,  respectively. 


To  prove  that  the  angles  ACB  and  ABB  are  supplemental. 

The  angle  ACB  is  half  the  central  angle  AOB  standing  on 
the  same  arc  ABB  (53). 

The  angle  ABB  is  half  the  central  angle  AOB  standing  on 
the  same  arc  ACB. 

Now  the  two  central  angles  AOB  together  make  up  a 
perigon.  Therefore  their  halves,  ACB  and  ADB,  together 
make  up  a  straight  angle ;  and  hence  are  supplemental. 

59.  Definition.  If  all  the  vertices  of  a  polygon  are  on  a 
circle,  the  polygon  is  said  to  be  inscribed  in  the  circle,  and 
the  circle  to  be  circumscribed  about  the  polygon. 

60.  Cor.  I .     The  opposite  angles  of  a  quadrangle  inscribed 

in  a  circle  are  supplemental. 

61.  Cor.  2.  An  extenor  angle  of  a  quadrangle  inscribed  in 
a  circle  equals  the  interior  opposite  angle. 

Ex.  1.  The  angle  in  a  segment  is  the  supplement  of  half  the 
central  angle  subtended  by  the  arc  of  the  segment. 

Ex.  2.     Similar  segments  of  circles  contain  equal  angles. 

Ex.  3.  Similar  segments  having  equal  chords  are  equal.  (Super- 
pose, use  56.) 


ANGLES  IN  SEGMENTS  191 

62.  Theorem  24.  If  the  opposite  angles  of  a  quad- 
rangle are  supplemental,  a  circle  may  be  circum- 
scribed  about  the  quadrangle. 

Let  the  quadrangle  ABCD  have  the  angles  ABC  and  ADC 
supplemental. 

To  prove  that  a  circle  may  be  described 
through  the  vertices  A,  B,  C,  D. 

Describe  a  circle  through  the  three 
points  A,  B,  c  (20). 

Suppose,  if  possible,  that  it  does  not 
pass  through  the  point  D.  Let  it  inter- 
sect AD  in  the  point  D'. 

The  angle  AD'C  is  supplemental  to  ABC  (58). 

Therefore  the  angles  ADC  and  AD'C  are  equal  (I.  51). 

But  this  equality  is  impossible  (I.  79). 

Therefore  the  supposition  is  false ;  hence  the  circle  pass- 
ing through  Ay  B,  c,  passes  also  through  D. 

EXERCISES 

1.  If  two  triangles  standing  on  the  same  base  and  at  the  same  side 
of  it  have  equal  vertical  angles,  then  the  circle  that  circumscribes  one 
triangle  will  circumscribe  the  other.     (Prove  similarly  to  62.) 

2.  If  two  chords  intersect  within  a  circle,  their  included  angle  is 
equal  to  half  the  sum  of  the  central  angles  standing  on  the  two  arcs 
intercepted  by  the  sides  of  thq  angle. 

3.  If  two  chords  when  extended  intersect  without  the  circle,  their 
included  angle  is  equal  to  half  the  difference  of  the  central  angles 
standing  on  the  two  arcs  intercepted  by  the  sides  of  the  angle. 

4.  The  bisectors  of  all  the  angles  in  a  given  segment  pass  through 
a  fixed  point.  The  bisectors  of  all  the  supplemental  angles  also  pass 
through  a  fixed  point. 

5.  The  lines  joining  adjacent  extremities  of  equal  chords  of  a  circle 
are  parallel. 

6.  The  lines  joining  the  extremities  of  parallel  chords  of  a  circle 
are  equal. 

7.  In  an  inscribed  quadrangle,  the  bisector  of  an  exterior  angle 
and  the  bisector  of  the  opposite  interior  angle  intersect  on  the  circle. 


192  PLANE  GEOMETRY  —  BOOK  III 

8.  If  a  parallelogram  is  circumscribable,  then  it  is  a  rectangle. 

9.  If  two  chords  of  a  circle  bisect  each  other,  then  their  intersec- 
tion is  the  center. 

10.  If  a  triangle  is  inscribed  in  a  circle,  then  the  sum  of  the  angles 
in  the  three  segments  exterior  to  the  triangle  is  equal  to  a  perigon. 

11.  If  a  quadrangle  is  inscribed  in  a  circle,  then  the  sum  of  the 
angles  in  the  four  segments  exterior  to  the  quadrangle  is  equal  to  six 
right  angles. 

12.  If  an  octagon  is  inscribed  in  a  circle,  then  the  sum  of  four  al- 
ternate angles  is  equal  to  the  sum  of  the  other  four. 

TANGENTS 

Conditions  for  Tangency 

The  following  fundamental  theorem  leads  up  to  the  defi- 
nition of  a  tangent  line  given  in  the  succeeding  article.  This 
theorem  and  the  next  establish  necessary  and  sufficient  con- 
ditions for  tangency,  and  lay  a  foundation  for  the  general 
theory,  and  for  the  problems  connected  with  it. 

63.  Theorem  25.  Of  all  the  straight  lines  that  can 
be  drawn  through  a  given  point  on  a  circle,  tJiere  is 
one,  and  one  only,  that  does  not  meet  the  circle 
again,  and  this  line  is  perpendicular  to  the  raMus 
drawn  to  the  given  point. 

Let  P  be  a  point  on  the  circle  whose  center  is  0.  Let 
NPQ  be  the  line  drawn  through  P  perpendicular  to  the 
radius  OP. 

To  prove  that  P  is  the  only  point  in 
which  the  line  NPQ  meets  the  circle. 

Take  any  point  Q  on  the  perpendicular 
NPQ\  and  draw  OQ. 

The  line  OQ  is  greater  than  OP  (1. 85). 

Therefore  Q  is  without  the  circle  (6).  ^ ^      ^ 

Hence  the  line  NPQ  meets  the  circle  in  one,  and  only  one, 
point. 


TANGENTS  193 

Next,  let  MPR  be  any  line  through  P  not  perpendicular  to 
the  radius  OP. 

To  prove  that  MPR  meets  the  circle 
in  a  second  point. 

Draw  OL  perpendicular  to  MPR. 
On  the  line  LR  lay  off  iQ  equal  to 
LP  ;  and  draw  0  Q. 

In   the   triangles   OLP,  OLQ,  the 
sides  PL,  LO  are  respectively  equal 
to  the  sides   QL,  LO,  and  the  included  angles  are  equal. 
Therefore  the  side  OQ  equals    'P  (I.  64). 

Hence  Q  is  a  point  on  the  circle  (6). 

Therefore  the  line  MPR  meets  the  circle  in  a  second  point  Q. 

64.  Definition.  The  straight  line  which  meets  the  circle 
in  a  given  point  and  does  not  meet  it  again  is  said  to 
touch  or  be  tangent  to  the  circle  at  that  point.  The  point 
is  called  the  point  of  contact  or  point  of  tangency. 

65.  Cor.  I.  Only  one  tangent  can  he  drawn  at  a  given 
'point  on  a  circle. 

66.  Cor.  2.  A  tangent  is  perpendicular  to  the  radius 
drawn  to  the  poiyit  of  contact;  and  a  line  passing  through  the 
extremity  of  a  radius,  not  perpendicular  to  it,  is  not  a  tangent. 

67.  Cor.  3.  The  center  lies  on  the  perpendicular  drawn  to 
any  tangent  at  the  point  of  contact. 

68.  Cor.  4.  The  perpendicular  from  the  center  to  any 
tangent  meets  it  in  the  point  of  contact. 

69.  Problem  2.  To  draw  a  tangent  to  a  given  circle 
at  a  given  point  on  the  circle.     [Use  63.] 

70.  Cor.  To  a  given  circle,  draw  a  tangent  that  shall  be 
parallel  to  a  given  line. 

Let  the  perpendicular  from  the  center  to  the  given  line  meet  the 
circle  in  P  and  Q.  The  tangents  drawn  at  P  and  Q  are  each  parallel 
to  the  given  line.     Prove. 


194 


PLANE  GEOMETRY — BOOK  III 


71.  Theorem  26.  A  given  line-  cuts,  touches,  or  does 
not  meet,  a  given  circle,  according  as  the  perpendicu- 
lar to  the  line  from  the  center  is  less  than,  equal  to, 
or  greater  than,  the  radius. 

Let  0  be  the  center  of  the  given  circle;  and  let  LN  be 
the  given  line.  Let  OM  be  the  perpendicular  from  the  center 
to  the  given  line. 

L  ^         ^L 


First,  let  OM  be  less  than  the  radius. 

To  prove  that  the  line  LMN  cuts  the  circle. 

On  the  line  LMN  lay  off  Af/>,  MN^  each  greater  than  the 
radius ;  and  draw  Oi,  ON. 

Since  OL  is  greater  than  LM  (1. 85)  OL  is  greater  than  the 
radius  and  the  point  L  is  without  the  circle  (6).  Similarly, 
the  point  N  is  without  the  circle.  But  since  OM  is  less  than 
the  radius,  the  point  M  is  within  the  circle  (6).  Thus  LMN  is 
a  secant. 

Next,  let  the  perpendicular  OM  be  equal  to  the  radius. 

Then  the  line  LMN  touches  the  circle  at  M  (63). 

Finally,  let  OM  be  greater  than  the  radius. 

To  prove  that  the  line  LMN  does  not  meet  the  circle. 

Let  L  be  any  point  on  the  line  LMN  except  the  point  M. 

Since  the  oblique  line  OL  is  greater  than  OM,  it  is  greater 
than  the  radius,  and  the  point  L  is  without  the  cirde. 

Again,  since  OM  is  greater  than  the  radius,  the  point  M  is 
without  the  circle.   Thus  LMN  is  entirely  without  the  circle. 

72.  Cor.  According  as  a  line  cuts,  touches,  or  does  not  meet, 
a  circle,  so  is  the  peiyendicular  to  the  line  from  the  center  less 
than,  equal  to,  or  greater  than,  the  radius. 


TANGENTS  195 

Tangents  from  External  Point 

73.  Theorem  27.  Through  a  given  external  point, 
two,  and  only  two,  lines  pass  that  touch  a  given  circle. 

Let  LMN  be  the  circle,  P  the  given  external  point. 

First,  to  prove  that 
two   tangents  to  the  ^^^^      ,,  ,'•'''       ~"^^^ 

circle    pass    through       '~~"^^~%''^ A^'^-^  \ 

the  point  P.  /    '^^"""^/    ^-Ar"^""""""--^  ^^ 

Take  the  center  0,  r/  V/  \      ^~""^^^^^:>>«^\ 

and  draw  OP.    On  OP  \  V\\      /       ^-z^:^^^^^^^'i 

as  diameter  describe  \  ^^^--''/QiS^^^^'^'^  ' 

the  circle  OMPN.  ,,-.-^^*<ClI,J<r*^ 

Since  0  is  within  ^^^^""^^      ^^-^^^         ^^^-^^ 

and  P  is  without  the 

circle  LMNj  hence  part  of  the  circle  OMPN  is  within  and 
part  is  without  the  circle  LMN  \  therefore  the  two  circles 
cut  in  at  least  two  points.  Moreover,  they  do  not  cut  in 
more  than  two  points  (22).  Let  M  and  N  be  the  two  points ; 
draw  PM,  PN,  OM. 

The  angle  OMP  is  a  right  angle  {^Q), 

Therefore  PM  is  tangent  at  the  point  M  (63). 

Similarly,  PN  is  tangent  at  N. 

Therefore  two  tangents  pass  through  the  given  point  P. 

Next,  to  prove  that  only  two  tangents  pass  through  P. 

Draw  any  other  line  PR  meeting  the  circle  LMN  in  at 
least  one  point. 

If  this  point  is  within  the  circle  OMPN  as  at  Q,  the  angle 
OQP  is  greater  than  the  angle  ONP,  and  hence  greater  than 
a  right  angle  (b^).    Therefore  PQR  is  not  tangent  at  Q  (63). 

Again,  if  any  line  PR^  meets  the  circle  LMN  in  a  point 
outside  the  circle  OMPN,  as  at  O',  the  angle  OQ^P  is  less 
than  OMP,  and  hence  less  than  a  right  angle  {^^)'  There- 
fore PQ^R^  is  not  a  tangent  at  Q'  (63). 

Hence  the  lines  PM,  PN  are  the  only  tangents  through  P. 


196  PLANE  GEOMETRY — BOOK  III 

74.  Problem  3.  To  draw  a  tangent  to  a  given  cir- 
cle from  a  given  external  point. 

Use  the  construction  and  proof  given  in  73.    Two  solutions. 

75.  Cor.  The  two  tangents  to  a  circle  from  an  external 
X)oint  are  equal,  and  make  equal  angles  with  the  line  joining 
that  point  to  the  center. 

Angle  of  Chord  and  Tangent 

76.  Theorem  28.  If  through  any  point  on  a  circle 
a  chord  and  a  tangent  are  drawn,  each  of  the  adjo/- 
cent  included  angles  is  equal  to  tlie  angle  in  the 
alternate  segment  of  the  circle  cut  off  by  the  chord. 

Through  the  point  P  on  the  circle  PQR,  let  the  chord  PQ 
and  the  tangent  MPN  be  drawn. 

First,  to  prove  that  the  acute  angle  y^ 

NPQ  is  equal  to  the  angle  in  the  alter-        / 
nate  segment  PRQ.  I 

Draw  the  diameter  PR ;  and  join  RQ.       I 

Since  Pff  is  a  diameter  and  PN  is        \ 
a  tangent,  the  angle  NPR  is  a  right  ^^  \/':>^ 

angle  (66). 

Therefore  the  angle  NPQ  is  the  complement  of  QPR, 

Now,  since  PQR  is  a  semicircle,  the  angle  PQi?  is  a  right 
angle  {hQ>). 

Therefore  the  angle  PRQ  is  the  complement  of  QPR 
(I.  129). 

Therefore  the  angles  NPQ  and  PRQ  are  equal,  being  com- 
plements of  the  same  angle  (I.  50). 

Hence  the  angle  NPQ  is  equal  to  any  angle  in  the  seg- 
ment PRQ  (54). 

Next,  to  prove  that  the  obtuse  angle  MPQ  is  equal  to  any 
angle  in  the  alternate  segment  PSQ. 

The  angle  MPQ  is  equal  to  the  supplement  of  NPQy  and 
the  angle  PSQ  is  equal  to  the  supplement  of  PRQ  (58). 


TANGENTS  197 

Hence  the  angles  MPQ,  FSQ,  being  equal  to  supplements 
of  equal  angles,  are  equal  (I.  51). 

Applications  of  Theorem  28 

77.  Problem  4.  On  a  given  line,  to  describe  a  seg- 
ment of  a  circle  containing  an  angle  equal  to  a 
given  angle. 

Let  AB  be  the  given  line,  and  C  the  given  angle. 

To  describe  on  AB  a  segment 
of  a  circle  containing  an  angle 
equal  to  C. 

Draw  AD,  making  the  angle 
BAD  equal  to  the  given  angle 
C  (I.  77). 

Draw  AO  perpendicular  to 
AD  ;  and  draw  MO  bisecting 
AB  at  right  angles.     Let  these  'n^^) 

lines  meet  in  0  ;  and  draw  OB. 

The  lines  OA  and  OB  are  equal  (I.  64). 

Therefore  the  circle  described  with  0  as  center,  and  with 
radius  OA,  passes  through  B.     Let  ABN  be  this  circle. 

The  segment  ABN,  alternate  to  the  angle  BAD,  is  the 
required  segment. 

Since  OAD  is  a  right  angle,  AD  touches  the  circle  (63). 

Therefore  the  angle  BAD  is  equal  to  the  angle  in  the 
alternate  segment  ABN  (76). 

Hence  the  angle  in  the  segment  ABN  is  equal  to  C. 

78i    Ex.     Consider  the  case  in  which  the  angle  O  is  a  right  angle. 

79.  Cor.  If  the  base  and  vertical  angle  of  a  triangle  are 
given,  the  locus  of  its  vertex  consists  of  the  arcs  of  the  two 
segments  described  on  the  base,  containing  an  angle  equal  to 
the  given  vertical  angle. 

Show  that  a  triangle  on  the  given  base  satisfies  the  requirements 
if  its  vertex  is  on  one  of  these  arcs,  and  not  otherwise. 


198 


PLANE  GEOMETRY  —  BOOK  III 


80.  Problem  5.  From  a  given  circle,  to  cut  off  a 
segment  containing  an  angle  equal  to  a  given  angle. 

Let  ABC  be  the  given  circle,  and  D  the  given  angle. 

To  cut  off  from  the  cir- 
cle ABC  a  segment  con-       Ay 
taining  an  angle  equal  to 
D. 

Through  any  point  B 
on  the  circle  draw  the 
tangent  EBF  (69).  Draw 
the  chord  BC  making  the 
angle  FBC  equal  to  the  given  angle  D  (I.  77). 

The  segment  BAG,  alternate  to  the  angle  FBC,  is  the 
segment  required. 

Since  BF  is  a  tangent,  the  angle  FBC  is  equal  to  the  angle 
in  the  alternate  segment  bIc  (76). 

Therefore  the  angle  in  the  segment  BAC  is  equal  to  the 
given  angle  D. 

Discussion.  How  many  solutions  are  there  to  this 
problem  ?  How  many  solutions  will  there  be  if  the  state- 
ment of  the  problem  is  modified  in  the  following  manner: 
Through  a  given  point  on  a  circle,  to  draw  a  line  that  shall 
cut  off  a  segment  containing  an  angle  equal  to  a  given  angle. 


EXERCISES 

1.  All  chords  of  a  circle  that  touch  a  concentric  circle  are  equal. 

2.  Through  a  given  point  draw  a  line  so  that  the  chord  intercepted 
by  a  given  circle  shall  be  equal  to  a  given  line.  [Use  ex.  1  and 
art.  74.  State  the  restrictions  on  the  data  when  the  point  is  within 
the  circle  ;  also  when  the  point  is  without  or  on  the  circle.  ] 

3.  The  part  of  any  tangent  intercepted  by  two  parallel  tangents 
subtends  a  right  angle  at  the  center  of  the  circle. 

4.  If  through  the  center  of  a  circle  two  perpendicular  lines  are 
drawn  to  meet  any  tangent,  then  the  tangents  drawn  from  the  two 
points  of  intersection  are  parallel. 


TWO  CIRCLES  199 

5.  To  draw  a  tangent  to  a  given  circle  making  a  given  angle  with 
a  given  line. 

6.  Any  chord  of  a  circle  bisects  the  angle  between  the  diameter 
through  one  extremity  and  the  perpendicular  from  it  on  the  tangent 
at  the  other. 

7.  Draw  a  circle  through  a  given  point  to  touch  a  given  line  at  a 
given  point.     [Use  3  and  67.] 

8.  If  a  quadrangle  is. circumscribed  about  a  circle,  the  sum  of  one 
pair  of  opposite  sides  is  equal  to  the  sum  of  the  other  pair. 

9.  If  a  convex  quadrangle  is  such  that  the  sum  of  one  pair  of  oppo- 
site sides  is  equal  to  the  sum  of  the  other  pair,  then  a  circle  may  be 
inscribed  in  it. 

TWO  CIRCLES 

81.  Definitions.  Two  circles  are  said  to  intersect  at  a 
point  where  they  meet  if  they  cross  each  other  at  this 
common  point. 

Two  circles  are  said  to  touch  at  a  point  where  they  meet 
if  they  do  not  cross  each  other  at  this  common  point ;  and 
this  point  is  called  the  point  of  contact. 


82.  The  line  passing  through  the  centers  of  two  circles 
is  called  their  central  line. 

Points  Common  to  Two  Circles 
Common  point  not  on  central  line, 

83.  Theorem  29.  If  two  circles  have  one  common 
-point,  not  on  their  central  line,  then  tlxey  have  a 
second  common  point;  and  the  circles  intersect  at 
each  of  these  two  points. 

Let  two  circles  whose  centers  are  0  and  O'  have  the 
common  point  P,  not  on  their  central  line. 

MCM.   ELEM,   GEOM.  14 


200  PLANE  GEOMETRY — BOOK  III 


First,  to  prove  that  they  have  a  second  common  point. 

Draw  PN  perpendicular  to  00',  and  prolong  it  to  Q  making 
NQ  equal  to  PJ^.     Draw  OP,  OQ,  O'P,  o'Q. 

By  equality  of  triangles  it  follows  that  OP  equals  OQ,  and 
O'P  equals  O'Q. 

Therefore  Q  is  a  point  on  each  of  the  circles  (6). 

Moreover,  there  is  no  third  common  point  (22). 

Next,  to  prove  that  the  circles  intersect  at  each  of  the 
points  P,  Q. 

Let  R,  S  be  any  two  points  on  the  circle  whose  center  is 
O',  situated  at  opposite  sides  of  the  point  P.  Draw  OR,  O'R, 
OS,  O'S. 

In  the  triangles  00' R  and  OO'P,  the  sides  00'  and  O'R  are 
respectively  equal  to  the  sides  00'  and  O'P,  and  the  included 
angle  OO'R  is  greater  than  OO'P. 

Therefore  the  third  side  07?  is  greater  than  the  third  side 
OP  (I.  91). 

Therefore  the  point  R  is  without  the  circle  whose  center 
is  0  (6). 

In  a  similar  way  it  is  proved  that  OS  is  less  than  OP. 

Therefore  the  point  S  is  within  the  circle  whose  center 
is  O. 

Now  R  and  S  are  aiiy  two  points  on  the  circle  whose 
center  is  o',  situated  at  opposite  sides  of  P. 

Hence  the  two  circles  cross  each  other  at  P. 

Similarly  it  can  be  proved  that  they  cross  at  Q. 


TWO   CIRCLES  201 

Common  point  on  central  line, 

84.  Theorem  30.  If  two  circles  have  one  common 
point,  situated  on  their  central  line,  then  they  have 
no  other  common  point;  and  the  circles  touch  at  this 
point. 

Let  0,  0'  be  the  centers  of  the  two  circles,  and  P  the 
common  point  on  the  central  line  00'. 

First,  to   prove   that  there  is                                        ^li 
no  other  common   point  on  the                               ^,'y'' 
central  line.  q 'o^' P 

Suppose,  if  possible,  that  Q  is 
another  point  on  the  line  00',  common  to  the  two  circles. 

Then  the  segment  PQ  is  a  diameter  of  each  circle.  Hence 
the  middle  point  of  PQ  is  the  center  of  each  circle;  which 
is  impossible  since  the  centers  0,  O'  do  not  coincide. 

Therefore  there  is  no  second  common  point  on  the  central 
line. 

Next,  to  prove  that  there  is  no  second  common  point  not 
on  the  central  line. 

Suppose,  if  possible,  that  R  is  another  common  point  not 
on  the  central  line.  Then  there  is  a  third  common  point 
22',  not  on  the  central  line  (83). 

Since  there  are  three  common  points,  the  two  circles  coin- 
cide throughout  (21). 

This  is  contrary  to  the  hypothesis ;  therefore  there  is  no 
common  point  not  on  the  central  line. 

Hence  P  is  the  only  common  point. 

Again,  to  prove  that  the  circles  touch  at  P. 

Let  R  be  any  other  point  on  the  circle  whose  center  is  0\ 
Draw  OR  and  O'iji. 

Since  O'R  equals  O'P,  therefore  the  sum  of  O'R  and  00'  is 
equal  to  OP. 

But  OR  is  less  than  the  sum  of  oo'  and  O'R  (L  87). 


202  PLANE  GEOMETRY  —  BOOK  III 

Therefore  OB  is  less  than  OP. 

Hence  the  point  R  is  within  the  circle  whose  center  is  0 
and  whose  radius  is  OP. 

Since  R  is  any  point  (other  than  P)  on  the  circle  whose 
center  is  O',  it  follows  that  the  circles  do  not  cross  at  their 
common  point  P.  Therefore  they  touch  at  this  point 
(81). 

JPoint  of  contact, 

85.  Theorem  31.  If  two  circles  touch,  then  their 
point  of  contact  is  on  the  central  line;  and  they  have 
no  other  common  point. 

Let  there  be  two  circles  touching  each  other  at  a  point. 


To  prove  that  the  point  of  contact  is  on  the  central  line ; 
and  that  the  circles  have  no  other  common  point. 

Suppose,  if  possible,  that  the  point  of  contact  is  not  on 
the  central  line. 

Then,  since  the  circles  have  a  common  point  not  on  the 
central  line,  they  intersect  at  this  point  (83). 

This  is  contrary  to  the  hypothesis  ;  hence  the  supposition 
made  is  false.  Therefore  the  point  of  contact  is  on  the 
central  line. 

Further,  since  the  two  circles  have  a  common  point  on 
their  central  line,  it  follows  that  they  have  no  other  common 
point  (84). 

86.  Cor.  I.  If  two  circles  touch  each  other  externally y  the 
line  joining  their  centers  is  equal  to  the  sum  of  their  radii. 


TWO  CIRCLES 


203 


87.  Cor.  2.  If  two  circles  touch,  one  being  internal  to  the 
other,  the  line  joining  their  centers  is  equal  to  the  difference  of 
their  radii. 

88.  Cor.  3.  If  two  circles  do  not  meet,  and  each  is  tvholly 
outside  the  other,  the  line  joining  their  centers  is  greater  than 
the  sum  of  their  radii. 

89.  Cor.  4.  If  two  circles  do  not  meet,  and  one  is  wholly 
inside  the  other,  the  line  joining  their  centers  is  less  than  the 
difference  of  their  radii. 

90.  Cor.  5.  If  two  circles  intersect,  the  line  joining  their 
centers  is  less  than  the  sum  of  their  radii,  and  greater  than  the 
differenxie  of  their  radii. 


91.  Note.  The  relation  of  these  live  cases  to  each  other  is  well 
shown  by  taking  first  the  case  in  which  each  circle  lies  wholly  outside 
the  other,  and  then  moving  one  center  toward  the  other,  as  succes- 
sively shown  in  the  figures. 

Ex.  1.  In  the  five  preceding  corollaries,  show  that  the  hypotheses 
are  exhaustive,  and  that  the  conclusions  are  mutually  exclusive  ;  then 
apply  the  rule  of  conversion  (I.  104)  to  prove  the  converse  of  each 
corollary.     JS.g.^ 

If  the  line  joining  the  centers  of  two  circles  is  less  than  the  differ- 
ence of  the  radii,  then  one  circle  is  wholly  within  the  other. 

Ex.  2.  Find  a  point  such  that  its  joins  to  two  given  points  may  be 
equal  respectively  to  two  given  lines.     (Intersection  of  loci.) 

Show  when  there  are  two  solutions,  when  only  one  solution,  and 
when  none.     Compare  I.  132. 

Ex.  3.  If  two  circles  touch,  they  have  a  common  tangent  at  the 
point  of  contact. 


204 


PLANE  GEOMETRY  —  BOOK  III 


Tangents  Common  to  Two  Circles 

92.  Problem  6.  To  draw  a  common  tangent  to  two 
given  circles. 

Let  0,  0'  be  the  centers  of  the  given  circles  QRS,  Q'b's', 
to  which  it  is  required  to  draw  a  common  tangent. 

First,  let  the  radii  be 
unequal;  and  let  the  cir- 
cle QRS  have  the  greater 
radius. 

Draw  any  radii  OS  and 
O's'.  From  the  greater 
OS  lay  off  sr  equal  to  the 
less  o's'.  With  O  as  center  and  OT  as  radius  describe 
a  circle  TPN.  From  the  point  O'  draw  a  tangent  o'P  to 
the  latter  circle.  Join  OP  and  prolong  it  to  meet  the  circle 
QRS  in  Q.     Draw  O'Q'  parallel  to  OQ,  and  join  QQ'. 

The  line  QQ'  touches  each  circle. 

[The  proof  is  left  to  the  student ;  also  the  construction  of  a  second 
common  tangent  by  nhe  same  method.  How  is  the  method  to  be 
modified  when  the  radii  are  equal  ?    Give  construction  and  proof.] 

If  the  circles  have  no  common  point  and  are  external  to 
each  other,  two  other  common  tangents  can  be  drawn  as 
follows : 

Let  OS  be  any  radius  of 
either  circle.  Prolong  it  so 
that  ST  equals  the  radius 
of  the  other  circle.  With 
O  as  center  and  OT  as 
radius,  draw  the  circle 
TPN.  From  the  point  O' 
draw  O'N  tangent  to  TPN. 

Draw  ON  cutting  the  circle  QRS  in  R.     Draw  the  radius  O'r' 
parallel  to  ORj  and  join  RR'.     Then  RR'  touches  each  circle. 

[The  proof  is  left  to  the  student ;  also  the  construction  of  a  second 
common  tangent.] 


P\ 


TWO   CIRCLES  205 

Discussion.  By  successively  moving  O'  toward  O,  as  in  91,  show 
when  there  are  four  common  tangents,  when  only  three,  when  only 
two,  when  only  one,  when  none. 

Definition.  A  line  touching  two  circles  is  said  to  be  a 
direct  or  a  transverse  common  tangent  according  as  the  two 
radii  drawn  to  the  points  of  contact  are  at  the  same  side  or 
at  opposite  sides  of  the  central  line. 

In  the  different  cases  just  mentioned,  how  many  of  the  common 
tangents  are  direct,  and  how  many  transverse  ? 

EXERCISES 

1.  Draw  a  tangent  to  one  given  circle  so  that  the  part  intercepted 
by  another  given  circle  shall  be  equal  to  a  given  line. 

2.  Draw  a  line  so  that  the  chords  intercepted  by  two  given  circles 
shall  be  respectively  equal  to  two  given  lines. 

3.  If  two  circles  touch,  and  through  the  point  of  contact  two  lines 
are  drawn  cutting  the  circles  again,  then  the  chords  joining  the  other 
intersections  are  parallel. 

4.  If  two  circles  touch,  and  through  the  point  of  contact  a  line  is 
drawn  to  cut  the  circles  again,  then  the  tangents  at  the  other  intersec- 
tions are  parallel ;  and  the  line  divides  the  two  circles  into  arcs  that 
are  respectively  similar, 

5.  If  two  circles  touch,  and  if  two  parallel  diameters  are  drawn, 
then  an  extremity  of  each  diameter  and  the  point  of  contact  are  in  the 
same  straight  line. 

6.  Describe  a  circle  through  a  given  point  and  touching  a  given 
circle  at  a  given  point.     (Determine  its  center  by  85  and  ex.  5.) 

7.  Describe  a  circle  to  touch  a  given  line,  and  touching  a  given 
circle  at  a  given  point.  (Draw  a  diameter  perpendicular  to  the  given 
line,  and  use  ex.  5.) 

8.  Describe  a  circle  to  touch  a  given  circle,  and  touching  a  given 
line  at  a  given  point. 

9.  The  two  circles  described  witU  two  sides  of  a  triangle  as  diam- 
eters intersect  on  the  third  side. 

10.  If  two  equal  circles  do  not  intersect,  show  how  to  draw  a  line 
so  that  its  extremities  and  points  of  trisection  may  be  on  the  two 
circles.  (When  are  there  two  solutions ;  only  one  solution ;  no 
solution  ?) 


206  PLANE  GEOMETRY  —  BOOK  III 

CONCURRENT  CHORDS 
Rectangle  of  segments  of  chord, 

93.  Theorem  32.  //  a  chord  of  a  circle  is  divided 
into  two  segments  hy  a  point  taken  either  in  the 
chord  or  in  its  prolongation,  the  rectangle  of  the  two 
segments  is  equivalent  to  the  difference  of  the  squares 
on  the  radius  and  on  the  line  joining  the  given  point 
to  the  center  of  the  circle. 

Let  the  chord  BC  of  the  circle  whose  center  is  0  be  divided 
at  the  point  A  into  the  two  segments  AB  and  AC. 

To  prove  that  the 
rectangle  of  AB  and 
AC  is  equivalent  to 
the  difference  of  the 
squares  on  OB  and 
OA. 

Draw  the  radius 
OC. 

Since  the  triangle  OBC  is  isosceles,  and  A  is  any  point  in 
the  base  or  its  prolongation,  therefore  the  rectangle  of  the 
segments  AB  and  AC  is  equivalent  to  the  difference  of  the 
squares  on  OB  and  OA  (II.  70). 

[Consider  the  special  case  in  which  BC  passes  through  0.] 

Several  concurrent  chords. 

94(a).  Cor.  i.  If  several  chords  pass  through  the  same 
point,  the  rectangle  of  the  segments  of  any  one  chord  is  equiva- 
lent to  the  rectangle  of  the  segments  of  any  other. 

Converse. 

94  (6).  Cor.  2.  If  two  given  lines  cut  each  other,  either 
internally  or  externally,  so  that  the  rectangle  of  the  segments 
of  one  is  equivalent  to  the  rectangle  of  the  segments  of  the  other, 
then  the  four  extremities  of  the  lines  lie  on  the  same  circle. 


CONCURRENT  CHORDS  207 

Chord  and  tangent, 

95.  Theorem  33.  If  a  chord  of  a  circle  is  prolonged 
to  any  point,  then  the  rectangle  of  the  segments  of 
the  chord  is  equivalent  to  the  square  on  the  tangent 
drawn  from  that  point  to  the  circle. 

Let  the  tangent  CP  and  the  chord  AB  meet  in  P. 

To  prove  that  the  rectangle 
of  AP  and  BP  is  equivalent  to 
the  square  on  CP. 

Take  the  center  0,  and  draw 
OA,  OB,  OC,  OP. 

The  rectangle  of  AP  and  BP 
is  equivalent  to  the  difference 
of  the  squares  on  OP  and  OB 
(93). 

But  this  difference  is  equivalent  to  the  difference  of  the 
squares  on  OP  and  OC,  which  is  equivalent  to  the  square 
on  CP  (II.  61).  Therefore  the  rectangle  of  AP  and  BP  is 
equivalent  to  the  square  on  CP. 

Ex.   Show  that  this  may  be  considered  a  special  case  of  94  (a). 

Converse, 

96.  Cor.  I.  If  the  rectangle  contained  by  the  segments  of  a 
chord  passing  through  an  external  point  is  equivalent  to  the 
square  on  the  line  joining  that  point  to  a  point  on  the  circle, 
then  this  line  touches  the  circle. 

Ex.   Draw  a  circle  through  two  given  points  to  touch  a  given  line. 

97.  Cor.  2.  If  several  circles  pass  through  the  same  two 
points,  then  the  tangents  drawn  to  them  Jrom  any  point  on  the 
prolongation  of  their  common  chord  are  all  equal. 

Ex.  If  two  circles  intersect,  and  if  a  tangent  is  drawn  to  each 
from  a  point  not  on  the  prolongation  of  their  common  chord,  then 
these  two  tangents  are  not  equal.  [Draw  a  secant  through  the  point 
and  one  of  the  intersections.     Use  95.] 


208  PLANE  GEOMETRY — BOOK  III 

EXERCISES 

1.  If  two  circles  intersect,  then  the  prolongation  of  their  common 
chord  bisects  their  common  tangent. 

2.  If  three  circles  intersect,  then  their  three  chords  of  intersection 
meet  in  a  point.     [Prove  indirectly,  using  94.] 

3.  If  two  circles  touch,  and  if  a  third  circle  intersects  them,  then 
the  tangent  at  the  point  of  contact  and  the  two  chords  of  intersection 
are  all  concurrent. 

4.  Draw  a  circle  through  two  given  points  to  touch  a  given  circle. 
[Determine  the  point  of  contact  by  means  of  ex.  3.  Two  solutions 
in  general.  Examine  the  case  in  which  one  of  the  given  points  is  on 
the  given  circle.     When  is  there  no  solution  ?] 

INSCRIPTION  AND   CIRCUMSCRIPTION 

98.  Definitions.  If  all  the  sides  of  a  polygon  touch  a 
circle  lying  within  the  polygon,  then  the  circle  is  said  to  be 
inscribed  in  the  polygon,  and  the  polygon  to  be  circum- 
scribed about  the  circle. 

If  a  circle  passes  through  all  the  vertices  of  a  polygon 
lying  within  the  circle,  then  the  polygon  is  said  to  be 
inscribed  in  the  circle,  and  the  circle  to  be  circinnscribed 
about  the  polygon.  A  circle  that  touches  one  side  of  a  tri- 
angle and  the  prolongations  of  the  other  two  is  said  to  be 
escribed  to  the  triangle. 

This  section  will  consist  chiefly  of  problems  relating  to 
the  inscription  and  circumscription  of  certain  regular  poly- 
gons to  a  circle,  and  of  a  circle  to  any  regular  polygon.  In 
the  particular  case  of  the  triangle,  however,  we  shall  not  be 
restricted  to  the  equilateral  triangle. 

Circles  and  Triangles 
Circumscribed  circle, 

99.  Problem  7.  To  circumscribe  a  circle  about  a 
given  triangle.     [Use  the  method  and  proof  of  20.] 

100.  Cor.  Every  triangle  has  oney  and  only  one,  circumr 
scribed  circle. 


INSCBIPTION  AND  CIItCUMSCRIPTION 


209 


Inscribed  circle, 

101.   Problem  8. 
triangle. 


To  inscribe   a  circle  in  a  given 


Let  ABC  be  the  triangle  in  which  it  is  required  to  inscribe 
a  circle. 

Bisect  any  two  of  the  internal 
angles,  say  B  and  C.  Let  the  bisectors 
BO  and  CO  meet  in  0.  Draw  OM,  ON, 
OP,  perpendicidar  respectively  to  the 
sides  BC,  CA,  AB. 

The  perpendiculars  OM,  ON,  and  OP 
are  all  equal  (I.  258). 

Therefore  the  circle  described  with  0  as  center  and  OM  as 
radius,  passes  through  the  points  M,  N,  P. 

The  circle  MNP,  so  described,  touches  the  sides  of  the 
triangle  at  M,  N,  P,  because  the  angles  0MB,  ONC,  and  OPA- 
are  right  angles  by  construction  (63). 

102.  Cor.  I.  Every  triangle  has  one,  and  only  one,  inscribed 
circle. 

Ex.  The  inscribed  and  circumscribed  circles  of  an  equilateral 
triangle  are  concentric. 

Escribed  circles, 

103.  Cor.  2.  To  escribe  a  cir- 
cle to  a  given  triangle. 

[Construct  and  prove  as  in  101.] 

104.  Cor.  3.  There  is  one,  and 
only  one,  circle  touching  a  given 
side  of  any  triangle  and  the  pro- 
longations of  the  other  two  sides; 
and  there  are  three,  and  only 
three,  escribed  circles  to  any  given 
triangle. 


210  PLANE   GEOMETRY  —  BOOK  III 

Definitions.  The  centers  of  the  circum  scribed,  the  inscribed, 
and  the  three  escribed  circles  are  called  respectively  the 
circum-center,  the  in-center,  and  the  three  ex-centers 
of  the  triangle. 

Ex.  1.  The  join  of  two  ex-centers  passes  through  a  vertex  of  the 
triangle  ;  the  join  of  the  third  ex-center  to  the  in-center  passes  through 
the  same  vertex ;  and  these  lines  are  perpendicular  to  each  other. 

Inscribed  triangle, 

105.  Problem  9.  In  a  given  circle,  to  inscribe  a 
triangle  equiangular  to  a  given  triangle,  and  having 
one  vertex  at  a  given  point  on  the  circle. 

Let  ABC  be  the  given  circle,  A^B^Cf  the  given  triangle,  and 
A  the  given  point. 


To  inscribe  in  ABC  a  triangle  equiangular  to  A'B'Cf,  and 
having  one  vertex  at  A. 

At  the  point  A  on  the  circle  draw  the  tangent  LAM. 
Draw  the  chord  AB  making  the  angle  LAB  equal  to  C'; 
draw  the  chord  AC  making  the  angle  MAC  equal  to  B' ;  and 
join  BC. 

Then  ABC  is  the  required  triangle. 

The  angle  LAB  equals  the  angle  ACB  in  the  alternate 
segment  of  the  circle  (76).    Therefore  the  angle  C  equals  C'. 

Similarly,  the  angle  B  equals  B'.  Hence  the  remaining 
angle  BAC  equals  the  remaining  angle  A'  (I.  130). 

Therefore  the  inscribed  triangle  ABC  is  equiangular  to 
the  triangle  A'b'c'. 


INSCRIPTION  AND  CIECUMSCRIPTION  211 

Discussion.  If  the  angle  LAB  had  been  made  equal  to  B',  and 
MAC  to  C,  another  triangle  answering  the  requirements  would  have 
been  obtained.  In  both  these  solutions  A  corresponds  to  A'.  Two 
more  solutions  can  be  obtained  in  which  A  corresponds  to  B',  and  two 
in  which  A  corresponds  to  C 

Show  that  the  number  of  solutions  would  be  reduced  to  two  if  we 
should  insert  in  the  statement  of  the  problem  the  additional  condi- 
tion: "the  vertex  to  which  the  given  point  is  to  correspond  being 
previously  assigned."  In  what  case  would  these  two  solutions  reduce 
to  one  9 

Ex.  Prove  that  all  triangles  inscribed  in  the  same  circle  and  equi- 
angular to  each  other,  are  equal. 

Circumscribed  triangle, 

106.  Problem  10.  About  a  given  circle  to  circum- 
scribe a  triangle  equiangular  to  a  given  triangle. 

[Use  72  ;  and  I.  126.] 

Ex.  1.  About  a  given  circle  to  circumscribe  a  triangle  equiangu- 
lar to  a  given  triangle,  one  of  the  three  points  of  contact  being 


[Rotate  the  given  triangle  so  that  one  of  the  sides  shall  be  parallel 
to  the  tangent  at  the  given  point  (I.  202),  and  then  use  72,  Three 
choices  of  correspondence ;  and  two  ways  of  rotating.  Discuss  as 
in  105.] 

Ex.  2.  Prove  that  all  triangles  circumscribed  about  the  same  circle 
and  equiangular  to  each  other,  are  equal. 

Principles  of  Inscription  and  Circumscription 

The  two  following  theorems  establish  the  general  prin- 
ciples that  will  be  used  in  the  problems  of  inscribing  and 
circumscribing  regular  polygons  to  given  circles.  These 
theorems  presuppose  the  division  of  the  circle  into  a  num- 
ber of  equal  arcs.  This  division  cannot,  however,  be 
actually  performed  by  the  constructions  of  elementary 
geometry,  except  in  the  case  of  certain  special  numbers, 
the  chief  of  which  are  shown  in  the  succeeding  group  of 
problems. 


212  PLANE  GEOMETRY  —  BOOK  III 

Principle  of  inscription, 

107.  Theorem  34.  If  a  circle  is  divided  into  a 
number  of  equal  arcs,  the  chords  of  these  arcs  form 
a  regular  inscribed  polygon. 

Let  the  circle  ABC  be  divided  into  a 
number  of  equal  arcs  at  the  points  A, 
B,  c,  '",  z,  and  let  the  chords  of  these 
arcs  be  AB,  BC,  CDy  •••,  ZA. 

To  prove  that  the  inscribed  polygon 
ABCD-"Z  is  regular. 

Since   the    arcs   AB^   BC,  •••  are  all 
equal,  their  chords  are  all  equal  (43).     Therefore  the  poly- 
gon is  equilateral. 

Again,  the  angles  of  the  polygon  are  equal  since  they  are 
angles  in  equal  segments  of  the  circle  (54).  Therefore  the 
polygon  is  equiangular. 

Hence,  by  definition,  the  polygon  is  regular. 

108.  Definitions.  If  the  extremities  of  a  broken  line 
coincide  with  the  extremities  of  an  arc,  and  if  all  its  ver- 
tices are  on  the  arc,  then  the  broken  line  is  said  to  be 
inscribed  in  the  arc. 

A  regular  broken  line  is  one  whose  sides  are  equal,  and 
whose  successive  angles  are  equal,  the  equal  angles  all  lying 
at  one  side  of  the  line. 

109.  Cor.  If  an  arc  is  divided  into  a  number  of  equal  arcs, 
then  their  chords  fomi  a  regular  inscribed  broken  line. 

Ex.  1.  Any  equilateral  polygon  inscribed  in  a  circle  is  also 
equiangular. 

Ex.  2.  If  an  equilateral  polygon  is  not  equiangular,  it  is  not  cir- 
cumscribable. 

Ex.  3.  In  any  equiangular  polygon  inscribed  in  a  circle,  each  side 
is  equal  to  the  next  but  one  ;  and  hence  an  inscribed  equiangular  poly- 
gon of  an  odd  number  of  sides  is  equilateral. 


INSCRIPTION^ AND   CIRCUMSCRIPTION  213 

JPrinciple  of  circumscription, 

110.  Theorem  35.  The  tangents  at  the  vertices  of 
an  inscribed  regular  polygon  form  a  circumscribed 
regular  polygon. 

Let  A,  B,  C,  D,  •••  be  the  vertices  of  an  inscribed  regular 
polygon ;  and  let  tangents  MAN,  NBP,  PCQ,  QBE,  •  •  •  be  drawn 
at  these  vertices. 

To  prove  that  the  circumscribed 
polygon  so  formed  is  regular. 

Find  the  center  O,  and  draw  OA, 
OB,  oc,  on,  •... 

Conceive  the  figure  turned  about  O 
until  OA  coincides  with  the  trace  of 
OB,  then  the  arc  AB  coincides  with 
its  equal   arc  BC,  and  the  line  OB  with  the  trace  of  OC. 
Similarly  OC  coincides  with  the  trace  of  OD,  and  so  on. 

Therefore  the  tangents  at  A,  B,  C,  •••,  being  perpendicular 
to  the  radii,  coincide  respectively  with  the  traces  of  the 
tangents  at  B,  C,  D,  ••• ;  and  hence  the  polygon  coincides 
with  its  trace. 

Therefore  the  angles  are  all  equal,  and  the  sides  are  all 
equal ;  hence  the  circumscribed  polygon  is  regular. 

Ex.  If  a  circumscribed  polygon  is  equiangular,  then  it  is  regular. 
[Prove  the  central  angles  AOB  and  BOC  equal,  etc.] 

Certain  Regular  Polygons 

111.  Division  of  the  Circle.  By  the  two  preceding  theorems 
the  inscription  and  circumscription  of  regular  polygons  have 
been  reduced  to  the  division  of  the  circle  into  a  given  num- 
ber of  equal  parts.  The  next  four  problems  with  their  corol- 
laries show  how  to  perform  the  actual  division  when  the 
given  number  belongs  to  one  of  the  four  following  series : 

2,  4,    8,  16,  32,  ... ;  5,  10,  20,    40,    80,  ... ; 

3,  6,  12,  24,  48,  ... ;  15,  30,  60,  120,  240,  ...  5 


214  PLANE  GEOMETRY -^  BOOK  III 

in  each  of  which  the  numbers  after  the  first  are  obtained  by 
successive  doubling.  These  numbers  will  for  convenience 
be  called  Euclid's  numbers,  as  the  problems  in  question 
were  first  systematically  treated  in  Euclid's  "  Elements  of 
Geometry." 

The  division  of  a  circle  into  two  equal  parts  is  easily  per- 
formed by  drawing  a  diameter.  As  this  division  does  not 
give  an  inscribed  polygon,  we  begin  with  the  second  number 
of  the  first  series. 


Four,  eight,  sixteen  •  •  •  sides, 

112.  Problem  11.     To  inscribe  a  square  in  a  given 
circle. 

[Draw  two  diameters  at  right  angles,  and  join  their  extremities.] 

113.  Cor.    I.     To  inscribe  a  regular  octagon  in  a  given 
circle. 

Outline.     Draw  two  diameters  at  right  angles  ;  bisect  the  four  arcs  ; 
draw  the  eight  chords.    Prove  by  107. 

114.  Cor.  2.     To  inscribe  regular  polygons  of  16, 32  •  •  •  sides. 

115.  Cor.  3.     To  circumscribe  regular  polygons  of  4,  8,  16, 
32,  ...  sides  (110). 

116.  Cor.  4.     In  any  given  arc  to  inscribe  a  regular  broken 
line  of  2,  4,  8,  16,  •••  sides. 

117.  Cor.  5.     TJte  inscribed  square  is  equivalent  to  double 
the  square  on  the  radius,  and  to  half  the  circumscribed  square. 


Three,  six,  twelve  •  •  •  sides, 

118.   Problem  12.     To    inscribe   an  equilateral    tri- 
angle in  a  given  circle. 

Let  ABC  be  the  circle  in  which  an  equilateral  triangle^  is 
to  be  inscribed. 


INSCRIPTION  AND   CIRCUMSCRIPTION  215 

Find  the  center  0,  and  take  any  point  P 
on  the  circle.  With  center  P  and  radius 
PO,  describe  an  arc  cutting  the  circle  at 
B  and  C.  Draw  PO  to  meet  the  circle 
again  in  A ;  and  draw  AB,  BC,  CA. 

The  inscribed  triangle  ABC  is  equi- 
lateral. 

The  triangles  BOP,  COP  are  each  equilateral  by  construc- 
tion; therefore  the  angles  BOP,  COP  are  each  equal  to  one 
third  of  a  straight  angle  (I.  129). 

Therefore  their  supplements  BOA,  CO  A  are  each  equal  to 
two  thirds  of  a  straight  angle;  hence  the  central  angles 
BOA,  COA,  and  BOC  are  all  equal ;  and  are  therefore  sub- 
tended by  equal  arcs  and  equal  chords. 

Therefore  the  inscribed  triangle  ABC  is  equilateral. 

Note.  This  problem  could  be  solved  as  a  special  case  of  problem  9, 
but  the  construction  would  not  be  so  simple  as  that  just  given. 

119.  Cor.   I.      To  inscribe  a  regular  hexagon. 

Outline.     Bisect  the  arcs  AB,  BC,  CA,  and  draw  the  six  chords. 

Ex.  Prove  that  the  side  of  a  regular  inscribed  hexagon  is  equal  to 
the  radius  of  the  circle.  Hence  give  another  method  of  inscribing  a 
regular  hexagon. 

120.  Cor.  2.     To  inscribe  regular  polygons  of  12,  24,  48,  ••• 

sides. 

121.  Cor.  3.  To  circumscribe  regular  polygons  of  3,  6,  12, 
24,  ...  sides  {110). 

Five,  ten,  twenty,  •  •  •  sides. 

122.  Problem  13.  In  a  given  circle,  to  inscribe  a 
regular  decagon. 

Let  ABC  be  the  given  circle  in  which  a  regular  decagon  is 
to  be  inscribed. 

MCM.   ELEM.  GEOM.  15 


216 


PLANE  GEOMETRY  —  BOOK  III 


Take  the  center  0,  and  draw 
any  radius  OA.  Divide  OA  at 
P,  so  that  the  rectangle  of  OA 
and  AP  is  equivalent  to  the 
square  on  OP  (II.  89). 

With  center  A  and  radius 
equal  to  OP,  describe  an  arc 
cutting  the  circle  in  B\  and 
join  AB. 

Then  AB  is  the  side  of  a 
regular  decagon  inscribed  in  ABC. 

To  prove  this,  draw  OB  and  PB ;  and  draw  a  circle  through 
the  points  0,  P,  B. 

The  square  on  AB  is  equivalent  to  the  rectangle  of  OA 
and  APy  by  construction. 

Therefore  the  line  AB  i^  tangent  to  the  circle  OPB  (96). 

Hence  the  angle  ABP  is  equal  to  the  angle  POB  in  the 
alternate  segment  (76). 

Therefore  the  whole  angle  ABO  is  equal  to  the  sum  of  the 
angles  POB  and  PBO,  and  therefore  equal  to  the  exterior 
angle  APB  (I.  128). 

Now,  the  angle  ABO  is  equal  to  the  angle  BAO  (I.  59). 

Therefore  the  angle  APB  equals  the  angle  BAO. 

Hence  the  opposite  sides  AB  and  PB  are  equal. 

Also,  AB  equals  OP  by  construction;  therefore  OP  equals 
PB'y  and  hence  the  opposite  angles  POB  and  PBO  are 
equal. 

Therefore  the  angle  ABO,  which  has  been  proved  equal  to 
the  sum  of  POB  and  PBO,  is  double  the  angle  POB. 

Hence  the  isosceles  triangle  OAB  has  each  of  the  angles 
at  the  base  equal  to  double  the  vertical  angle  AOB. 

Therefore  the  angle  AOB  \^  equal  to  one  fifth  of  the  sura 
of  the  three  angles  of  the  triangle  OAB,  that  is,  equal  to 
one  fifth  of  a  straight  angle,  and  hence  equal  to  one  tenth  of 
a  perigon.     Thus  the  arc  ^jB  is  a  tenth  part  of  the  circle  (31). 


INSCRIPTION  AND   CIRCUMSCRIPTION  217 

Hence  the  circle  can  be  divided  into  ten  arcs  equal  to  AB ; 
and  the  chords  of  these  arcs  will  form  a  regular  inscribed 
decagon  (107). 

123.  Cor.  I.  To  inscribe  a  regular  pentagon  in  a  given 
circle.  [Join  alternate  vertices  of  an  inscribed  regular 
decagon  (107).] 

124.  Cor.  2.     To  inscribe  regular  polygons  of  20,  40,  ••• 

sides. 

125.  Cor.  3.  To  circumscribe  regular  polygons  of  5,  10, 
20,  •••  sides. 

126.  Cor.  4.  Show  that  the  central  angle  subtended  by  the 
side  of  an  inscribed  regidar  decagon  is  two  fifths  of  a  right 
angle.     Show  how  to  divide  a  right  angle  into  five  equal  parts. 

127.  Cor.  5.  On  a  given  line,  to  construct  a  regular 
decagon. 

[In  any  circle  inscribe  a  regular  decagon.  At  the  extremities  of 
the  given  line,  make  angles  equal  to  the  angle  of  the  regular  decagon, 
and  so  on.] 

Ex.  1.  The  triangle  OAB  is  an  isosceles  triangle  having  each  angle 
at  the  base  equal  to  double  the  vertical  angle. 

Ex.  2.  The  triangle  FOB  is  an  isosceles  triangle  having  each  angle 
at  the  base  equal  to  one  third  of  the  vertical  angle. 

Ex.  3.  If  C  is  the  second  point  of  intersection  of  the  circles  OPB 
and  ABC,  then  BC  equals  AB.  li  D  is  the  next  vertex,  and  if  the 
circle  OPB  cuts  the  radius  OD  again  in  Q,  then  OPBCQ  is  a  regular 
pentagon. 

Ex.  4.  Prove  that  BP  prolonged  passes  through  a  vertex  I  of  the 
decagon  ;  and  that  the  difference  of  jB/and  BA  is  equal  to  the  radius. 

Ex.  5.  Show  that  OCP  is  an  isosceles  triangle,  and  that  the  rec- 
tangle of  OA  and  PA  is  equivalent  to  the  difference  of  the  squares  on 
AC  and  OC ;  that  is  to  say,  the  difference  of  the  squares  on  the  sides 
of  the  inscribed  pentagon  and  decagon  is  equivalent  to  the  square  on 
the  radius. 

Ex.  6.  If  a  regular  pentagon  and  a  regular  decagon  are  inscribed  in 
the  same  circle,  then  the  apothem  of  the  pentagon  equals  half  the  sum 
of  "a  side  of  the  decagon  and  a  radius  of  the  circle.  [BK  bisects  AP ; 
use  I.  72.] 


218  PLANE  GEOMETRY— BOOK  III 

Fifteen,  thirty,  sixty,  .  .  .  sides, 

128.  Problem  14.  To  inscribe  a  regular  qulndeca- 
gon  in  a  given  circle. 

Let  ABC  he  the  circle  in  which  it  is  required  to  inscribe 
a  regular  polygon  of  fifteen  sides. 

Find  the  center  0,  and  draw  a  radius  OA.     Divide  OA  at 
the  point  JV,  so  that  the  rectangle  of 
OA  and  ON  is  equivalent  to  the  square         /^^^    ^^\ 
on  AN  (II.  89).  /  \ 

With  A  as  center,  and  AN  and  AO  sls    I  q  \ 

radii,  describe  arcs  cutting  the  circle  in    I  A\  I 

B  and  C;  and  draw  BC.  \  j   \  \  ,    / 

Then  the  chord  BC  is  a  side  of  a        ^v.^x^r^;^^^'^ 
regular  inscribed  quindecagon.  '^ 

The  chord  ^B  is  a  side  of  a  regular  inscribed  decagon 
(122);  and  the  chord  ^C  is  a  side  of  a  regular  inscribed 
hexagon  (119). 

Therefore,  if  the  whole  circle  were  divided  into  thirty 
equal  parts,  the  arc  AB  would  contain  three  of  them,  and 
the  arc  AC  would  contain  five.  Therefore,  the  arc  BC  would 
contain  two  of  these  parts,  and  is  hence  one  fifteenth  of  the 
whole  circle.  Thus  the  circle  is  divisible  into  fifteen  arcs,  each 
equal  to  the  arc  BC]  therefore  the  polygon  formed  by  the 
chords  of  these  arcs  is  a  regular  quindecagon  (107).  Hence 
the  chord  5(7  is  a  side  of  a  regular  inscribed  quindecagon. 

129.  Cor.  To  inscribe  regular  polygons  of  30,  60  sides; 
and  to  circurmcnhe  ■  regular  polygons  of  15,  30,  60  sides. 

Ex.  1.  To  divide  a  right  angle  into  fifteen  equal  parts. 

Ex.  2.  To  divide  the  angle  of  an  equilateral  triangle  into  five  equal 
parts. 

Ex.  3.  To  trisect  the  angle  of  a  regular  pentagon. 

Note  on  the  numbers  of  Gauss.  The  four  preceding  problems 
with  their  corollaries  have  shown  how  to  divide  the  circle  into  n  equal 
parts,  if  n  is  any  one  of  Euclid's  numbers  (111).     From  the  time  of 


INSCRIPTION  AND   CIRCUMSCRIPTION  219 

Euclid  no  essential  advance  was  made  in  the  problem  of  dividing  the 
circle  until  the  year  1790,  when  Gauss*  proved  the  possibility  of  divid- 
ing the  circle  into  n  equal  parts,  if  n  is  any  prime  number  that  exceeds 
a  power  of  2  by  unity.  The  first  four  numbers  that  satisfy  this  con- 
dition are  8,  5,  17,  257.  Gauss  further  proved  that  the  division  can 
be  performed  if  n  is  the  product  of  any  two  or  more  different  numbers 
of  this  series  ;  the  first  four  numbers  that  satisfy  this  condition  are  15, 
51,  85,  255.  Gauss  gave  the  complete  analysis  for  the  case  of  17 
parts,  and  proved  that  the  problem  can  be  reduced  to  simpler  ones 
that  depend  ultimately  on  the  postulates  of  construction  ;  but  the 
method  of  proof  is  beyond  the  range  of  elementary  geometry. 

Inscribed  and  Circumscribed  Circles 

The  following  theorem  and  its  corollaries  furnish  the 
basis  for  the  two  succeeding  problems,  which  relate  to  the 
construction  of  the  inscribed  and  circumscribed  circles  of 
any  given  regular  polygon. 

Concurrence  of  angle-bisectors, 

130.  Theorem  36.  The  lines  that  bisect  the  angles 
of  any  regular  polygon  all  meet  in  a  point. 

Outline.  Let  A,  B,  C,  D  be  consecutive  angles  of  any 
regular  polygon.  Bisect  the  angles  A  and  B\  and  prove 
that  the  bisectors  meet  at  that  side  of  the  line,  AB,  at  which 
the  polygon  itself  is  (I.  124).  Let  the  bisectors  meet  at  the 
point  0,  and  draw  00.  Prove  that  00  bisects  the  angle  0. 
(This  is  done  by  proving  that  the  angle  OOB  equals  OAB, 
which  equals  half  of  the  angle  A,  and  hence  equals  half  O.) 
Prove  similarly  that  OD  bisects  the  angle  D ;  and  so  on. 

131.  Cor.  I.  In  any  regular  polygon,  the  segments  of  the 
angle-bisedors  intercepted  between  the  vertices  and  the  point  of 
concurrence  are  all  equal. 

132.  Cor.  2.  Inany  regular  polygon,  the  perpendiculars  from 
the  intersection  of  the  angle-bisectors  to  the  sides  are  all  equal. 

*  "  Disquisitiones  Arithraeticae,"  published  in  1801. 


220  PLANE  GEOMETRY  —  BOOK  III 

Circumscribed  circle, 

133.  Problem  15.  To  circumscribe  a  circle  about  a 
given  regular  polygon.     (Use  130,  131.) 

Inscribed  circle, 

134.  Problem  16.  To  inscribe  a  circle  in  a  given 
regular  polygon.     (Use  130,  132.) 

135.  Cor.  Tlie  inscribed  and  circumscribed  circles  of  a 
regular  polygon  are  concentric. 

Ex.  1.  If  two  regular  polygons  are  equal,  then  their  inscribed 
circles  are  equal,  and  so  are  their  circumscribed  circles. 

Ex.  2.  If  two  regular  polygons  of  the  same  number  of  sides  are 
inscribed  in  equal  circles,  then  the  two  polygons  are  equal. 

136.  Definitions.  The  common  center  of  the  inscribed 
and  circumscribed  circles  of  a  regular  polygon  is  called*  the 
center  of  the  regular  polygon.  The  angle  at  the  center 
subtended  by  any  side  of  the  polygon  is  called  the  central 
angle  of  the  regular  polygon. 

In  a  regular  polygon,  a  line  joining  the  center  to  any 
vertex  is  called  a  radius,  and  a  perpendicular  from  the 
center  to  any  of  the  sides  is  called  an  apothem.  Thus  a 
radius  of  a  regular  polygon  is  a  radius  of  its  circumscribed 
circle,  and  an  apothem  is  a  radius  of  its  inscribed  circle. 

Ex.  1.  If  any  two  regular  polygons  have  the  same  number  of  sides, 
then  their  central  angles  are  equal. 

Ex.  2.  If  two  regular  polygons  have  the  same  number  of  sides,  and 
if  the  radius  of  one  is  greater  than  the  radius  of  the  other,  then  the 
apothem  of  the  first  is  greater  than  the  apothem  of  tlie  second,  the  side 
of  the  first  is  greater  than  the  side  of  the  second,  and  the  surface  of 
the  first  is  greater  than  the  surface  of  the  second. 

[Place  the  polygons  with  their  centers  in  coincidence,  and  so  that 
each  radius  of  the  first  may  fall  on  a  radius  of  the  second  ;  then  prove 
that  each  side  of  the  first  is  parallel  to  a  side  of  the  second ;  etc.] 

Ex.  3.  If  two  regular  polygons  have  the  same  number  of  sides,  then 
the  following  pairs  of  magnitudes  are  in  the  same  order  of  size  : 

The  bases  (/>,  &');  the  radii  (r,  r');  the  apothems  (a,  a');  the  sur- 
faces (s,  s');  the  perimeters  (p,  ;>'). 


INSCRIPTION  AND   CIRCUMSCRIPTION  221 

Equivalent  rectangle, 

137.  Theorem  37.  A  polygon  circumscribed  about  a 
circle  is  equivalent  to  the  rectangle  contained  by  the 
perimeter  and  half  the  radius  of  the  circle. 

Ex.  A  regular  polygon  is  equivalent  to  the  rectangle  contained  by 
the  perimeter  and  half  the  apothem. 

MUTUALLY   EQUILATERAL    POLYGONS 

138.  Theorem  38.  If  two  circles  are  equal,  and  if 
two  mutually  equilateral  polygons  are  inscribed  i?i 
them,  then  the  polygons  are  equal. 

Outline.  Compare  the  respective  triangles  whose  vertices 
are  at  the  centers  and  whose  bases  are  corresponding  sides 
of  the  polygons. 

139.  Theorem  39.  //  two  mutually  equilateral 
polygons  are  each  circumscribable  by  a  circle,  then 
the  circles  are  equal,  and  tJie  polygons  are  equal. 

Outline.  Suppose  the  radii  unequal.  Compare  the  respec- 
tive central  angles  subtended  by  corresponding  sides;  see 
ex.  42  at  end  of  Book  I.     Eeduce  to  absurdity. 

140.  Cor.  If  in  two  semicircles  are  inscribed  two  broken 
lines,  and  if  the  segments  of  one  broken  line  are  respectively 
equal  to  those  of  the  other,  taken  in  order,  then  the  semicircles 
are  equal,  and  the  two  figures  are  superposable.  [Prove  as  in 
139.] 

EXERCISES 

1.  Two  regular  polygons  of  the  same  number  of  sides  circumscribed 
about  equal  circles  are  equal. 

2.  The  center  of  the  inscribed  circle  of  a  triangle  is  the  ortho- 
center  of  the  triangle  formed  by  the  centers  of  the  escribed  circles. 

3.  Show  how  to  cut  off  the  corners  of  an  equilateral  triangle  so 
as  to  leave  a  regular  hexagon  ;  also  of  a  square  to  leave  a  regular 
octagon. 

4.  If  a  parallelogram  is  circumscribed  to  a  circle,  then  it  is  a 
rhombus. 


222  PLANE  GEOMETRY — BOOK  III 

5.  Show  that  it  is  possible  to  trisect  the  central  angle  of  a  regular 
n-gon,  when  n  is  any  one  of  the  first  or  third  series  of  Euclid's  num- 
bers (111). 

6.  In  the  same  cases  show  that  it  is  possible  to  trisect  the  interior 
or  exterior  angle  of  the  regular  w-gon. 

7.  Show  that  it  is  possible  to  divide  a  right  angle  into  n  equal 
parts,  when  n  is  any  of  Euclid's  numbers. 

8.  Show  that  it  is  possible  to  divide  the  angle  of  an  equilateral  tri- 
angle into  n  equal  parts,  when  n  is  any  number  belonging  to  the  first 
three  series  of  Euclid's  numbers. 

Note  on  exs.  5-8.  The  general  problem  to  trisect  a  given  arbitrary 
angle  is  one  of  the  famous  problems  of  antiquity,  and  has  never  been 
solved  by  methods  permitted  in  elementary  geometry.  Modern 
mathematicians  have  demonstrated  that  this  general  problem  cannot 
be  analyzed  into  simpler  ones  that  require  only  the  drawing  of  straight 
lines  and  circles.*  Thus  the  construction  cannot  be  performed  by 
means  of  only  a  pair  of  compasses  and  an  unmarked  straightedge. 
Several  general  solutions  are  known  which  overstep  these  limitations 
to  a  greater  or  less  degree.  One  of  the  simplest  employs  the  sliding 
motion  of  a  straightedge  on  which  two  points  are  marked.  There 
are,  however,  certain  special  angles  that  can  be  trisected  by  the 
methods  of  elementary  geometry.     (See  exs.  5-8  above.) 

The  still  more  general  problem  of  dividing  a  given  arbitrary  angle 
into  n  equal  parts  can  be  solved  only  when  n  is  one  of  the  first  series 
of  Euclid's  numbers  (111  ;  I.  73)  ;  but  in  the  case  of  certain  special 
angles  the  problem  can  be  solved  for  some  other  values  of  n  (exs.  7,  8). 

MAXIMA  AND  MINIMA  t 

141.  Certain  principles  of  maxima  and  minima  relating 
to  triangles  were  considered  in  Book  II.  92-108.  Similar 
principles  can  now  be  extended  to  polygons  in  general,  sub- 
ject to  certain  given  conditions.  The  theorems  here  con- 
sidered fall  into  five  groups  according  to  the  nature  of  the 
assigned  conditions. 

*  See  Klein's  "Vortrage  iiber  ausgewahlte  Fragen  der  Elementar 
Geometrie."     (Translated  by  Professors  Beman  and  Smith.) 

t  This  topic  is  discussed  here  on  account  of  its  intimate  connection 
with  the  properties  of  the  circle,  and  of  inscribed  and  circumscribed 
polygons.     It  may,  however,  be  postponed  without  inconvenience. 


MAXIMA  AND  MINIMA  223 

Given   Sides 

This  group  of  two  theorems  with  their  corollaries  will 
show  how  to  make  the  surface  of  a  polygon  a  maximum 
subject  to  various  assigned  conditions  relating  to  the  magni- 
tude of  the  sides.  In  each  case  the  additional  condition  is 
to  be  proved  both  necessary  and  sufficient  for  a  maximum  ; 
and  accordingly  each  theorem  is  accompanied  by  its  con- 
verse (II.  93). 

Greatest  polygon  with  one  arbitrary  side, 

142.  Theorem  40.  Ainong  the  polygons  that  have 
all  the  sides  hut  one  equal  respectively  to  given  lines 
taken  in  order,  ariy  one  that  is  a  maxiwium  is  cir- 
cumscrihable  hy  a  semicircle  having  the  undetermined 
side  as  diameter. 

Let  the  polygon  ABCDEF  be  a  maximum  subject  to  the 
condition  that  the  sides  AB,  BC,  CD,  BE,  EFsltq  respectively 
equal  to  given  lines  taken  in 
order. 

To  prove  that  the  semi- 
circle described  on  AF  as 
diameter  passes  through  all 
the  points  B,  C,  D,  E. 

Suppose,  if  possible,  that 
the  semicircle  does  not  pass  through  C ;  and  draw  CA,  CF. 

Then  the  angle  ACF  is  not  a  right  angle  (55,  bOt). 

Hence,  by  rotating  the  figures  ABC  and  FEDC  about  the 
point  C  until  ACF  becomes  a  right  angle,  the  triangle  ACF 
could  be  increased  (II.  94);  and  therefore  the  whole  polygon 
ABCDEF  could  be  increased  without  changing  any  of  the 
given  sides.  This  is  contrary  to  the  hypothesis  that 
ABCDEF  is  a  maximum  under  the   given  conditions. 

Hence  the  semicircle  described  on  ^i^  passes  through  the 
point  C.     Similarly  it  passes  through  the  other  vertices. 


224 


PLANE  GEOMETRY  —  BOOK  III 


143.  Cor.  Among  the  x>olygons  that  have  all  the  sides  but 
one  equal  resj)ectively  to  given  lines  taken  in  order,  any  poly- 
gon that  is  circumscribable  by  a  semicircle  having  the  undeter- 
mined side  as  diameter,  is  a  maximum. 

For  all  polygons  that  satisfy  the  given  conditions,  and  the 
further  condition  of  being  circumscribable  by  a  semicircle 
having  the  undetermined  side  as  diameter,  are  equal  (140), 
and  are  therefore  equal  to  any  one  that  is  a  maximum  (142). 

Ex.    Show  how  to  enunciate  II.  94  so  as  to  make  it  a  particular 
of  143. 


Greatest  polygon  with  nil  the  sides  given, 

144.  Theorem  41.  Among  the  polygons  that  have 
their  sides  equal  respectively  to  given  lines  taken  in 
order,  any  polygon  that  is  cvrcumscrvbable  by  a  circle 


IS  a  maximum. 


Let  ABCD  and  a'b'c'd'  be  two  polygons  that  satisfy  the 
conditions  of  having  their  sides  equal  respectively  to  given 
lines  taken  in  order,  and  let  the  former  be  circumscribable 
and  the  latter  not. 


First  to  prove  that  ABCD  is  greater  than  A'b'c'd'. 

Draw  the  diameter  AP  ;  and  join  CP,  DP.  On  C'd',  which 
equals  CD,  construct  a  triangle  c'd'p'  equal  to  the  triangle 
CDP'j  and  draw  A'p'. 

The  circle  whose  diameter  is  A'P*  does  not  pass  through  all  the  points 
B',  C",  D'  (hyp.).     Use  143  and  add  ;  then  subtract  the  equal  triangles. 


MAXIMA   AND  MINIMA  225 

Next,  to  prove  that  ABC  Diss,  Tnaximum  under  the  given 
conditions. 

The  polygon  ABCD  is  superposable  on  any  other  polygon 
that  satisfies  the  given  conditions  and  the  further  con- 
dition of  being  circumscribable  (139);  and  it  has  just 
been  proved  greater  than  any  polygon  that  satisfies  the 
given  conditions  without  satisfying  the  further  condition. 
Therefore  the  polygon  ABCD  is  a  maximum  under  the  given 
conditions. 

145.  Cor.  I.  Among  the  polygons  that  have  their  sides 
respectively  equal  to  the  given  lines  taken  in  order,  any  polygon 
that  is  a  maximum  is  circumscribable.     (Indirect  proof.) 

Equilateral  n-gon  ivUh  given  side, 

146.  Cor.  2.  Of  all  equilateral  polygons  having  a  given 
side  and  a  given  number  of  sides,  one  that  is  regidar  is  a 
maximum. 

Among  the  polygons  that  satisfy  the  given  conditions, 
one  that  is  equilateral  and  equiangular  is  circumscribable ; 
and  one  that  is  equilateral  and  not  equiangular  is  not  cir- 
cumscribable (109,  ex.  2) ;  hence  one  that  is  equilateral  and 
equiangular  is  a  maximum  (144). 

147.  Cor.  3.  Of  all  equilateral  polygons  haviiig  a  given 
side  and  a  given  number  of  sides,  any  polygon  that  is  a 
maximum  is  regular.     (Use  145.) 

Given  Perimeter 

The  following  theorems  show  how  to  make  the  surface  of 
a  polygon  a  maximum,  when  the  perimeter  is  given,  and 
when  another  assigned  condition  is  fulfilled. 

The  n-gon  of  greatest  surface, 

148.  Theorem  42.  Among  the  polygons  that  have  a 
given  perimeter  and  a  given  number  of  sides,  one 
that  is  a  majoimuin  is  regular. 


226  PLANE  GEOMETRY— BOOK  III 

Let  the  polygon  ABCD  . . .  be  a  maximum,  subject  to  the 
conditions  of  having  a  given  perimeter  and  a  given  number 
of  sides. 

First,  to  prove  that  ABCD  ...  is  an  ^^'0^^\ 

equilateral  polygon.  a^ 

Suppose,  if  possible,  that  the  two 
adjacent  sides  AB  and  BC  are  not 
equal. 

On  ^C  as  base  construct  an  isosceles 
triangle  B^AC  having  the  sum  of  the 
sides  b'a  and  B'c  equal  to  the  sum  of  J5^  and  BC. 

The  isosceles  triangle  B*AC  is  greater  than  the  isoperi- 
metric  triangle  BAC  (II.  101). 

Therefore  the  polygon  AB'CD  ...  is  isoperimetric  with, 
and  greater  than,  the  polygon  ABCD  . . .]  but  this  is  im- 
possible, since  ABCD  .  .  .  is  one  of  the  greatest  of  the 
isoperimetric  set,  by  hypothesis. 

Hence  the  supposition  fails,  and  the  adjacent  sides  J  B  and 
BC  are  equal. 

Similarly  all  the  sides  are  equal.  Therefore  the  polygon 
ABCD  ...  is  equilateral. 

Next,  to  prove  that  the  equilateral  polygon  ABCD  ...  is  a 
regular  polygon. 

Since  ABCD  . . .  is  an  equilateral  polygon  having  a  given 
perimeter  (that  is  to  say,  a  given  side)  and  a  given  number 
of  sides,  and  since  it  is  a  maximum,  hence  it  is  a  regular 
polygon  (147). 

149.  Cor.  Among  the  polygons  that  have  a  given  perimeter 
and  a  given  number  of  sides,  one  that  is  regular  is  a  maxi- 
mum. 

For  all  polygons  that  satisfy  the  given  conditions  and  the 
additional  condition  of  being  regular  are  equal,  and  are 
hence  equal  to  any  one  that  is  a  maximum. 


MAXIMA  AND  MINIMA 


227 


Regular  polygon  of  greatest  surface, 

150.  Theorem  43.  Of  two  isoperimetric  regular 
polygons,  that  which  has  the  greater  number  of  sides 
has  the  greater  surface. 

Let  the  polygons  P  and  Q  have  equal  perimeters,  and  let  P 
have  one  side  more  than  Q. 

C 


To  prove  that  P  is  greater  than  Q. 

On  one  of  the  sides  of  Q  take  any  point  D. 

The  figure  ADBC  md^j  be  regarded  as  an  irregular  polygon 
having  the  same  number  of  sides  as  P ;  and  hence  ADBC  is 
less  than  the  isoperimetric  regular  polygon  P  (148,  149). 

Therefore  Q  is  less  than  P. 

Given  Surface 

The  follovring  two  theorems  show  how  to  make  the  pe- 
rimeter of  a  polygon  a  minimum  when  the  surface  is  given 
and  when  another  assigned  condition  is  fulfilled. 


The  n-gon  of  least  perimeter, 

151.  Theorem  44.  Among  the  polygons  having  a 
given  surface,  and  a  given  numher  of  sides,  one 
that  is  regular  has  a  minimuin  perimeter. 

Let  P  and  Q  be  two  polygons,  each  having  the  given  sur- 
face, and  the  given  number  of  sides ;  and  let  P  be  regular, 
and  Q  not  regular. 


228 


PLANE  GEOMETRY  —  B00:K  III 


First,  to  prove  that  P  has  a  less  perimeter  than  Q  has. 

Let  i?  be  a  regular  polygon  having  the  same  number  of 
sides,  and  the  same  perimeter,  as  Q  has. 

Of  the  two  isoperimetric  polygons  Q  and  /?,  the  latter, 
being  regular,  has  the  greater  surface.  Thus  the  surface  of 
R  is  greater  than  that  of  Q,  and  hence  greater  than  the 
surface  of  P. 

Now  the  regular  polygons  P  and  R  have  the  same  number 
of  sides,  hence  the  one  that  has  the  less  surface  has  the  less 
perimeter  (136,  ex.  3).  Thus  the  perimeter  of  P  is  less  than 
that  of  i?,  and  hence  less  than  that  of  Q. 

Next,  to  prove  that  the  perimeter  of  P  is  a  minimum  under 
the  given  conditions. 

The  perimeter  of  P  is  equal  to  that  of  any  regular  polygon 
satisfying  the  given  conditions  (136,  ex.  3),  and  has  been 
proved  less  than  the  perimeter  of  any  irregular  polygon 
satisfying  the  given  conditions.  Hence  the  perimeter  of  P 
is  a  minimum  under  the  given  conditions. 

152.  Cor.  Among  the  poIygo7i8  having  a  given  number  of 
sides  and  a  given  surface^  one  that  has  a  minimum  perimeter 
is  regular.     [Use  indirect  proof.] 

Regular  polygon  of  least  peHmeter, 

153.  Theorem  45.  If  any  two  regular  polygons 
have  equivalent  surfa/^es,  then  the  one  that  has  the 
greater  numher  of  sides  has  the  less  perimeter. 

Let  P  and  Q  represent  any  two  equivalent  regular  poly- 
gons, and  let  Q  have  more  sides  than  P  has. 


MAXIMA  AND  MINIMA 


229 


To  prove  that  the  perimeter  of  Q  is  less  than  that  of  P. 

Let  i?  be  a  regular  polygon  having  the  same  number  of 
sides  as  F,  and  the  same  perimeter  as  Q. 

Then,  of  the  two  isoperimetric  regular  polygons  Q  and  7?, 
the  former  has  the  greater  number  of  sides,  therefore  it  has 
the  greater  surface  (150) ;  thus  the  surface  of  R  is  less  than 
that  of  Q,  and  hence  less  than  that  of  P. 

Now  the  regular  polygons  P  and  R  have  the  same  number 
of  sides,  hence  the  one  that  has  the  less  surface  has  the  less 
perimeter  (136,  ex.  3) ;  thus  the  perimeter  of  R  is  less  than 
that  of  P.     Hence  the  perimeter  of  Q  is  less  than  that  of  P. 

Inscribed  in  Given  Circle 

154.  In  this  group  of  theorems  one  of  the  stated  con- 
ditions is  that  the  polygons  in  question  are  inscribed  in  a 
given  circle.  The  theorems  show  how  to  make  the  surface, 
or  the  perimeter,  greatest  when  the  polygon  is  subject  to 
assigned  conditions.  In  each  of  the  two  divisions  of  this 
group  we  begin  as  usual  with  the  case  of  the  triangle  and 
proceed  from  it  to  the  general  polygon. 


MAXIMUM    SURFACE 

Greatest  triangle  in  segment, 

155.  Theorem  46.  Ainong  the  triangles  inscribed 
in  the  same  segment  of  a  circle  having  their  bases 
coincident  with  the  chord,  the  triangle  that  is 
isosceles  is  the  maximum. 

[The  tangent  at  the  mid-point  of  the  arc  is  parallel  to  the  chord.  ] 


230  PLANE  GEOMETRY — BOOK  III 

156.  Cor.  I.  Among  the  Mangles  inscribed  in  the  same 
segment  of  a  circle  having  their  bases  coincident  with  the  chordj 
the  triangle  that  is  the  maximum  is  isosceles, 

157.  Cor.  2.  Of  all  triangles  inscribed  in  the  same  cirdey 
one  that  is  a  maximum  is  equilateral.     [Indirect  proof.] 

158.  Cor.  3.  Of  all  triangles  inscribed  in  the  same  circle, 
one  that  is  equilateral  is  a  maximum.   [Prove  as  in  149.] 

Greatest  n-gon  in  circle, 

159.  Theorem  47.  Among  the  polygons  having  a 
given  number  of  sides  and  inscribed  in  a  given  circle, 
one  that  is  a  myO^vimum  is  regular.    [Prove  as  in  157.] 

160.  Cor.  Among  the  polygons  having  a  given  number  of 
sides  and  inscribed  in  a  given  circle,  one  that  is  regular  is 
a  maximum.     [Prove  as  in  138  and  149.] 

Greatest  regular  polygon  in  circle, 

161.  Theorem  48.  If  any  two  regular  polygons  are 
inscribed  in  the  same  circle,  then  the  one  that  has 
the  greater  number  of  sides  ha^  the  greater  surface. 

Outline.  Let  there  be  two  regular  inscribed  polygons  P 
and  Q,  and  let  Q  have  one  more  side  than  P  has. 

To  prove  that  the  surface  of  Q  is  greater  than  that  of  P. 

Let  AB  be  one  side  of  P.  On  the  arc  AB  take  any  point 
M\  and  draw  MA,  MB.  Let  the  new  polygon  made  up  of 
the  polygon  P  and  the  triangle  MAB  be  denoted  by  P'. 

Then  p'  and  Q  have  the  same  number  of  sides.  Show  that 
Q  is  greater  than  P'  (160) ;  and  hence  greater  than  P. 

MAXIMUM    perimeter 

Triangle  in  given  segment, 

162.  Theorem  49.  Among  the  triangles  inscribed 
in  a  given  segment  of  a  circle  having  their  bases 
coincident  with  the  chord,  the  triangle  that  is  isosce- 
les has  the  maximum  perimeter. 


MAXIMA  AND  MINIMA  231 

Outline.  Let  ABC  and  AB'C  be  triangles  inscribed  in  the 
same  segment ;  and  let  ABC  be  isosceles,  having  the  side  AB 
equal  to  BC. 

To  prove  that  the  perimeter  of  ABC  is  greater  than  that 
oiAB'c. 

Prolong  AB  to  D  so  that  BD  equals  BC  ;  and  prolong  AB' 
to  Z)' so  that  5 'Z)' equals  5' C. 

Prove  that  the  angles  ADC  and  ad'c  are  equal,  being 
halves  of  equal  angles ;  and  hence  that  the  four  points  C, 
A,  B\  D,  are  on  the  same  circle,  whose  center  is  B.  Then 
prove  AD  greater  than  AD' ;  etc. 

163.  Cor.  I.  Among  the  triangles  inscribed  in  a  given  seg- 
ment of  a  circle  having  their  bases  coincident  with  the  chord, 
the  triangle  that  has  the  maximum  perimeter  is  isosceles. 

Triangle  in  given  circle, 

164.  Cor.  2.  Among  the  triangles  inscribed  in  a  given  circle, 
one  that  has  a  maximum  perimeter  is  equilateral. 

Prove  as  in  157,  using  163. 

165.  Cor.  3.  Among  the  triangles  inscribed  in  a  given  circle, 
one  that  is  equilateral  has  a  maximum  perimeter.     [See  149.] 

The  n-gon  of  greatest  perimeter, 

166.  Theorem  50.  A-mong  the  polygons  having  a 
given  number  of  sides  and  inscribed  in  a  given  circle, 
one  that  has  a  majcimum  perimeter  is  regular, 
[See  164.] 

167.  Cor.  Among  the  polygons  having  a  given  number  of 
sides  and  inscribed  in  a  given  circle,  one  that  is  regular  has 
a  maximum  perimeter.     [Prove  as  in  149,  165.] 

Regular  polygon  of  greatest  perimeter, 

168.  Theorem  51.  If  any  two  regular  polygons  are 
inscribed  in  the  same  circle,  then  the  one  that  has  the 
greater  number  of  sides  has  the  greater  pervmeter. 

MCM.  ELEM.  GEOM. — 16 


232  PLANE  GEOMETRY  —  BOOK  III 

Outline.  Let  there  be  two  regular  inscribed  polygons 
P  and  Q,  and  let  Q  have  one  side  more  than  P  has. 

To  prove  that  the  perimeter  of  Q  is  greater  than  that  of  P. 

Let  AB  \yQ  one  side  of  P.  On  the  arc  AB  take  any 
point  M\  and  draw  MA,  MB.  Let  the  new  polygon,  made 
up  of  the  polygon  P  and  the  triangle  MABy  be  denoted  by  P'. 

Show  that  the  perimeter  of  the  regular  polygon  Q  is 
greater  than  that  of  P'  (167);  and  that  the  perimeter  of 
P'  is  greater  than  that  of  P. 

Circumscribed  about  Given  Circle 

169.  In  this  group  of  theorems  one  of  the  stated  con- 
ditions is  that  the  polygons  in  question  are  circumscribed 
about  a  given  circle;  and  it  is  shown  how  to  make  the 
surface,  or  the  perimeter,  a  minimum  when  the  polygon  is 
subject  to  assigned  conditions.  The  first  division  of  this 
group  relates  to  minimum  surface,  the  second  to  minimum 
perimeter.  In  each  case  the  additional  condition  should  be 
proved  to  be  both  necessary  and  sufficient  for  a  minimum. 

MINIMUM    SURFACE 

Least  triangle  about  sector, 

170.  Theorem  52.     If  at  any  point  on  the  arc  of  a 

given  sector  a  tangent  is  drawn  to  meet  the  two 
ra^ii  prolonged,  then  the  triangle  so  formed  is  the 
minim,um^  when  the  tangent  is  drawn  at  the  mid- 
point of  the  arc. 

Outline.  Let  OAB  be  the  given  sector,  M  the  mid-point 
of  the  arc  AB,  and  m'  any  other  point  of  the  arc.  Suppose 
3/'  to  be  taken  on  the  half  arc  MB.  Let  the  tangent  at  M 
meet  the  prolongations  of  the  radii  OA  and  OB  in  the  points 
L  and  N.  Let  the  tangent  at  3/'  meet  the  same  prolonga- 
tions in  the  points  L'  and  N'  \  and  let  it  intersect  the  pre- 
ceding tangent  in  the  point  I. 

Prove  that  the  triangle  NIN'  is  less  than  iiz.';  etc. 


MAXIMA   AND  MINIMA  233 

One  point  of  contact  arbitrary, 

171.  Theorem  53.  If  all  the  points  of  contact  but 
one  are  assigned,  at  which  the  sides  of  a  circum- 
scribed polygon  touch  a  given  circle,  and  if  all  the 
points  of  contact  have  an  assigned  order  on  the  circle, 
then  the  circumscribed  polygon  is  least  when  the  arbi- 
trary point  of  contact  bisects  the  arc  whose  cjctremi- 
ties  are  at  the  two  points  of  contact  adjacent  to  that 
arbitrary  point. 

Outline.  Let  0  be  the  center  of  the  given  circle.  Among 
the  assigned  points  of  contact  let  P  and  R  be  the  two  which 
are  to  be  adjacent  to  the  unassigned  point,  and  let  Q  be  any 
position  of  the  unassigned  point  on  the  arc  PR.  Let  the 
tangents  at  P  and  R  meet  the  tangent  at  Q  in  the  points  L 
and  N  respectively.  Draw  OL  and  ON^  meeting  the  circle 
in  A  and  B  respectively. 

Show  that  the  angle  LON  is  half  the  angle  POR,  and  is 
hence  of  constant  magnitude,  whatever  be  the  position  of  Q 
on  the  arc  PR\  that  the  sector  AOB  is  of  constant  magni- 
tude; and  hence  that  the  triangle  LON  is  least  when  Q  is 
at  the  mid-point  of  the  arc  AB  (170)  ;  that  the  pentagon 
OPLNR  is  double  the  triangle  LON,  and  is  therefore  least 
w\\Qn  Q  is  at  the  mid-point  of  the  arc  AB  ;  show  that  Q  is 
then  at  the  mid-point  of  the  arc  PR ;  and  draw  conclusion. 

The  n-ffon  of  least  surface, 

172.  Theorem  54.  Among  all  the  polygons  of  a 
given  number  of  sides  circumscribed  about  a  given 
circle,  one  that  is  a  minimum  is  regular. 

Outline.  By  indirect  proof,  using  171 ,  show  that  any  point 
of  contact  bisects  the  arc  lying  between  the  two  adjacent 
points  of  contact ;  and  then  show  that  the  polygon  is  regular. 

173.  Cor.  Among  all  the  polygons  of  a  given  number  of 
sides  circumscribed  about  a  given  circle,  one  that  is  regidar 
is  a  minimum.     [Prove  as  in  149.     See  140,  ex.  1.] 


234  PLANE  GEOMETRY — BOOK  III 

Regular  polygon  of  least  surface, 

174.  Theorem  oo.  If  any  two  regular  polygons  are 
circumscribed  about  a  given  circle,  the  one  that  has 
the  greater  number  of  sides  has  the  less  surface. 

Outline.  Let  there  be  two  regular  circumscribed  poly- 
gons P  and  Q,  and  let  Q  have  one  more  side  than  P  has. 

To  prove  that  the  surface  of  Q  is  less  than  that  of  P. 

Let  T  be  any  one  of  the  vei-tices  of  P,  and  let  A  and  B  be 
the  points  of  contact  of  the  tangents  from  T.  At  any  point 
M  of  the  arc  AB  draw  the  tangent  LMN,  meeting  TA  in  L 
and  TB  in  N. 

The  tangent  LMN  cuts  off  a  triangle  LTN  from  the  poly- 
gon P,  leaving  a  circumscribed  polygon  which  has  one  side 
more  than  P  has.  Let  the  new  polygon  be  denoted  by  P'. 
Then  P'  and  Q  have  the  same  number  of  sides. 

Show  that  the  regular  polygon  Q  is  less  than  P'  (172, 173), 
and  hence  less  than  P. 

MINIMUM    PERIMETER 

175.  Theorem  56.  Among  the  polygons  of  a  given 
number  of  sides  circumscribed  about  a  given  circle, 
one  that  has  a  minimum  perimeter  is  regular. 

Outline.  The  surface  of  a  circumscribed  polygon  is  equiva- 
lent to  half  the  rectangle  of  the  perimeter  and  the  radius  of 
the  circle  (137) ;  hence,  when  the  perimeter  is  a  minimum, 
the  surface  is  a  minimum.     Then  apply  172. 

176.  Cor.  Among  the  polygons  of  a  given  number  of  sides 
circumscribed  about  a  given  circle,  one  that  is  regular  has  a 
minimum  perimeter.     [Prove  as  in  173.] 

Regular  polygon  of  least  peHmeter, 

177.  Theorem  57.  If  any  two  regular  polygons 
are  circumscribed  about  a  given  circle,  the  one 
thojt  has  the  greater  number  of  sides  hOjS  the  less 
perimeter.     [Prove  as  in  174,  and  use  176.] 


LOCUS  PBOBLEMS  235 

LOCUS  PROBLEMS 
Equal  tangents  to  two  circles, 

178.  Problem  17.  To  fund  the  locus  of  a  point 
from  which  the  tangents  drawn  to  two  given  circles 
are  equal. 

Outline.  If  the  circles  intersect,  show  by  means  of  97 
and  ex.  that  the  locus  is  the  extension  of  the  common  chord. 

If  the  circles  do  not  intersect,  let  P  be  a  point  from  which 
the  tangents  PT  and  PT'  drawn  to  the  circles  are  equal; 
and  let  0  and  O'  be  the  centers.  Using  II.  61,  prove  that 
the  difference  of  the  squares  on  OP  and  o'P  is  equivalent  to 
the  difference  of  the  squares  on  the  radii  OT  and  O'r'. 
Thus  the  finding  of  the  locus  is  reduced  to  II.  91. 

Consider  also  the  case  in  which  the  circles  touch. 

179.  Definition.  The  line  which  is  the  locus  of  a  point 
from  which  the  tangents  to  two  given  circles  are  equal  is 
called  the  radical  axis  of  the  two  circles. 

If  the  two  circles  intersect,  then  their  common  chord  is 
the  radical  axis. 

180.  Cor.  I.  Tlie  radical  axis  of  tivo  circles  is  perpendicu- 
lar to  their  central  line,  and  divides  it  either  internally  or 
externally  so  that  the  difference  of  the  squares  on  the  segments 
is  eqidvalent  to  the  difference  of  the  squares  on  the  radii. 

Equal  tangents  to  three  circles, 

181.  Cor.  2.  To  find  'a  point  from  which  the  tangents  to 
three  given  circles  are  equal. 

[Intersection  of  loci.     When  is  there  no  solution  ?] 

182.  Cor.  3.  Tlie  three  radical  axes  of  three  circles  meet  in 
a  point. 

Use  178,  181.  Consider  separately  the  case  in  which  there  are 
three  chords  of  intersection  that  divide  each  other  internally  ;  and  use 
ex.  2  following  Art.  97. 


236  PLANE  GEOMETRY  —  BOOK  III 

183.  Definition.  The  point  of  concurrence  of  the  three 
radical  axes  of  three  circles  is  called  the  radical  center  of 
the  three  circles. 


Circles  intersecting  orthogonally, 

184.  Definition.  If  two  circles  intersect,  and  if  the  two 
tangents  drawn  at  one  of  their  common  points  are  at  right 
angles,  then  the  circles  are  said  to  intersect  orthogonally. 

Ex.  If  two  circles  intersect  orthogonally  at  one  of  their  common 
points,  then  they  intersect  orthogonally  at  the  other  common  point. 

185.  Theorem  58.  //  two  circles  intersect  orthog- 
onally, then  the  tangent  drawn  to  one  of  them 
at  a  common  point  parses  through  the  center  of  the 
other. 

186.  Cor.  To  find  the  locus  of  the  center  of  a  circle  that 
cuts  a  given  circle  orthogonally  at  a  given  j^oint. 

Ex.  To  describe  a  circle  through  a  given  point,  and  cutting  a  given 
circle  orthogonally  at  a  given  point. 

[Determine  the  center  of  the  required  circle  by  the  intersection  of 
two  loci.] 

187.  Theorem  59.  //  two  tangents  are  drawn  to 
a  circle  from  an  external  point,  then  the  circle 
described  with  the  point  as  center,  and  either  of 
the  tangents  as  radius,  cuts  the  given  circle  orthog- 
onally. 

188.  Cor.  To  find  the  locus  of  the  center  of  a  circle  that 
cuts  two  given  circles  orthogonally. 

Show  that  the  required  locus  is  the  radical  axis. 
Ex.    To  describe  a  circle  cutting  three  given  circles  orthogonally. 
[Show  that  there  is  no  solution  if  each  circle  intersects  the  other 
two.] 


LOCUS  PROBLEMS  237 

Points  in  a  Triangle 

189.  There  is  a  class  of  locus  problems  in  which  the  base 
and  vertical  angle  of  a  triangle  are  given,  to  find  the  loci  of 
certain  important  points  connected  with  the  triangle.  It 
has  been  shown  in  79  that  the  locus  of  the  vertex  of  a  tri- 
angle which  has  a  given  fixed  base  and  a  given  vertical 
angle  consists  of  the  arcs  of  two  segments  described  on  the 
base  (one  on  each  side  of  it)  containing  an  angle  equal  to 
the  given  vertical  angle.  The  problems  and  corollaries  in 
190-193  are  reducible  to  the  one  just  mentioned. 

190.  Problem  18.  To  find  the  locus  of  the  ortho- 
centers  of  all  the  triangles  that  have  a  given  fixed 
base  and  a  given  vertical  angle. 

Outline.  Let  P  be  the  ortho-center  of  any  triangle  ABC 
standing  on  the  given  base  BC,  and  having  the  vertical 
angle  A  equal  to  the  given  angle. 

Show  that  the  triangle  BPC,  standing  on  the  given  base, 
has  its  vertical  angle  P  equal  to  the  supplement  of  the  given 
angle ;  hence  that  the  locus  of  P  consists  of  the  two  arcs 
obtained  by  describing  on  B  C  two  segments,  each  containing 
an  angle  equal  to  the  supplement  of  the  given  angle. 

191.  Cor.  I.  To  find  the  locus  of  the  inrcenters  of  all  the 
triangles  that  have  a  given  fixed  base  and  a  given  vertical  angle. 

Outline.  If  P  is  the  in-center,  show  that  the  angle  BPC  is 
equal  to  the  sum  of  the  given  vertical  angle  and  half  its 
supplement,  and  apply  79  as  before. 

192.  Cor.  2.  With  the  same  data,  find  the  locus  of  each  of 
the  three  ex-centers. 

Outline.  If  P  is  any  of  these  centers,  show  that  the 
angle  BPC  can  be  expressed  in  terms  of  .4.  In  two  of  the 
cases  BPC  equals  half  A ;  and  in  the  third  case  BPC  equals 
the  complement  of  half  A. 

Ex.     With  the  same  data,  show  that  the  circum-center  is  fixed. 


238  PLANE  GEOMETRY — BOOK  III 

193.  Problem  19.  To  find  the  locus  of  the  median- 
centers  of  all  the  triangles  that  have  a  given  fixed 
hase  and  a  given  vertical  angle. 

Outline.  Through  the  median-center  P  draw  parallels  to 
the  two  sides  AB  and  AC,  meeting  the  base  in  Q  and  R. 

Prove  that  Q  and  R  trisect  the  base.  Show  that  the 
triangle  PQR,  standing  on  the  middle  segment,  has  a  given 
fixed  base  and  a  given  vertical  angle.     Then  apply  79. 

194.  Problem  20.  Find  tJie  locus  of  the  vertices  of 
all  the  triangles  that  have  a  given  fixed  base,  and 
the  sum  of  tJw  squares  on  the  two  sides  equivalent 
to  a  given  square. 

Outline.  Show  from  II.  67,  76-78,  that  the  median  AD 
can  be  constructed  from  the  data.  Show  that  the  locus  of 
^  is  a  circle  with  D  as  center,  and  with  radius  equal  to  the 
line  AD  ^o  found. 

Subtended  Angles 

195.  Definitions.  The  angle  subtended  at  a  given  point 
by  a  given  line  is  the  angle  included  by  the  two  lines  drawn 
from  the  given  point  to  the  extremities  of  the  given  line. 

The  angle  subtended  at  a  given  point  by  a  given  circle 
is  the  angle  included  between  the  two  tangents  drawn  from 
the  given  point  to  the  circle. 

196.  Problem  21.  To  find  the  locus  of  a  point  at 
which  the  angle  subtended  by  a  given  line  is  equal 
to  a  given  angle.     [Compare  79.] 

Ex.  1.  To  find  on  a  given  indefinite  line  a  point  at  which  the  angle 
subtended  by  a  given  line-segment  shall  be  equal  to  a  given  angle. 
[When  are  there  two  solutions,  when  only  one,  and  when  none  ?] 
Ex.  2.  To  find  on  a  given  indefinite  line  a  point  at  which  a  given 
line-segment  shall  subtend  the  greatest  angle.  [See  ex.  following 
Art.  96.  Observe  the  change  in  the  subtended  angle  as  the  point  takes 
different  positions  on  the  line.  Is  there  any  position  at  which  the  sub- 
tended angle  is  least  ?] 


LOCUS  PROBLEMS  239 

Ex.  3.  To  find  on  a  given  circle  a  point  at  which  a  given  line  shall 
subtend  a  given  angle. 

[Discuss  the  solution  as  in  ex.  1.  Show  that  in  certain  cases  there 
is  no  solution  unless  the  given  angle  is  restricted  in  magnitude.] 

Ex.  4.  To  find  on  a  given  circle  the  points  at  which  the  angle 
subtended  by  a  given  line  is  a  maximum  or  minimum.     [97,  ex.  4.] 

197.  Problem  22.  To  find  the  locus  of  a  point  at 
which  the  angle  subtended  by  a  given  circle  shall  be 
equal  to  a  given  angle. 

Outline.  Show  that  the  locus  is  a  concentric  circle,  and 
that  its  radius  may  be  constructed  as  follows:  Construct 
a  right  triangle  having  one  side  equal  to  the  radius  of 
the  given  circle,  and  the  adjacent  acute  angle  equal  to  the 
complement  of  half  the  given  angle;  then  the  hypotenuse  is 
the  required  radius. 

Ex.  In  the  four  exercises  of  196,  replace  the  given  line-segment  by 
a  given  circle,  and  show  that  similar  solutions  can  be  obtained. 

Intersection  of  Loci 

198.  Each  of  the  following  constructions  is  a  combination 
of  two  locus  problems  already  solved. 

Ex.   1.     To  construct  a  triangle,  being  given  : 

(a)  its  base,  vertical  angle,  and  altitude  (79,  I.  255)  ; 
(5)  its  base,  vertical  angle,  and  difference  of  squares  on 

sides  (79,  II.  91)  ; 
(c)  its  base,  vertical  angle,  and  sum  of  squares  on  sides ; 
{d)  its  base,  altitude,  and  sum  of  squares  on  sides  (194)  ; 
(e)  its  base,  vertical  angle,  and  one  side  ;  the  base  being 
given  in  position  as  well  as  magnitude. 
Ex.  2.     To  construct  a  quadrangle,  being  given  two  opposite  angles, 
and  three  sides ;  the  order  in  which  the  five  parts  are  to  be  taken 
being  specified. 

Outline.  Let  the  sides  AB,  BC^  CD  be  given  ;  and  also  the  angles 
B  and  D.  First  construct  the  triangle  ABC  (I.  133)  ;  and  then  the 
triangle  ACD  (ex.  1,  e). 

[Examine  the  case  in  which  one  of  the  given  angles  is  convex.  Show 
that  there  is  always  a  solution  when  the  sum  of  the  two  given  angles 
is  less  than  a  perigon.     Show  that  there  is  only  one  solution.] 


240  PLANE  GEOMETRY  —  BOOK  III 


EXERCISES 

1.  If  from  any  point  on  a  given  circle  a  line  is  drawn  equal  and 
parallel  to  a  given  line,  then  the  locus  of  the  other  extremity  consists 
of  two  circles  each  equal  to  the  given  circle. 

[Take  the  given  line  as  "  line  of  translation  "  (I.  200),  and  translate 
the  center  and  any  radius,  thus  reducing  the  locus  problem  to  a 
previous  one  (7).] 

2.  If  from  any  point  on  a  given  circle  a  line  is  drawn  to  a  given 
point,  and  if  this  line  is  turned  about  the  given  point  through  a  given 
angle,  then  the  locus  of  the  other  extremity  of  the  line  so  turned  con- 
sists of  two  circles  each  equal  to  the  given  one. 

[Take  the  given  angle  as  "angle  of  rotation"  (I.  202),  and  rotate 
the  circle  and  any  radius  about  the  given  point.] 

3.  To  describe  three  circles  of  given  radii  to  touch  each  other 
externally. 

4.  To  describe  three  circles  of  given  radii  to  touch  each  other  so 
that  two  may  be  within  the  third. 

5.  In  an  equilateral  triangle  the  radius  of  each  of  the  escribed 
circles  is  equal  to  the  altitude  ;  and  the  radii  of  the  circumscribed  and 
inscribed  circles  are  respectively  equal  to  two  thirds,  and  one  third  of 
the  altitude. 

6.  If  two  chords  of  a  circle  cut  at  right  angles,  then  the  sum  of  either 
pair  of  opposite  arcs  is  equal  to  a  semicircle. 

[Through  an  extremity  of  one  chord,  draw  a  chord  parallel  to  the 
other,  and  join  its  extremity  to  the  other  extremity  of  the  first  chord.] 

7.  If  two  chords  of  a  circle  cut  at  right  angles,  then  the  sum  of  the 
squares  on  the  four  segments  is  constant,  and  equivalent  to  the  square 
on  the  diameter. 

8.  If  any  chord  of  a  given  circle  passes  through  a  fixed  point,  then 
the  rectangle  of  the  segments  of  the  chord  is  constant. 

9.  If  through  a  fixed  point  within  a  given  circle  any  two  chords  are 
drawn  at  right  angles,  then  the  sum  of  the  squares  on  the  two  chords 

is  constant  (exs.  7,  8). 

10.    If  each  of  two  equal  circles  has  its  center  on  the  circumference 

of  the  other,  then  the  square  on  their  common  chord  is  equivalent  to 
three  times  the  square  on  the  radius. 


EXERCISES  241 

11.  If  two  given  circles  have  external  contact ;  show  how  to  draw 
a  line  through  the  point  of  contact  so  that  the  whole  intercepted  part 
may  be  equal  to  a  given  line. 

[Form  an  isosceles  triangle  whose  base  is  the  given  line  and  each  of 
whose  other  sides  equals  the  sum  of  the  radii.  Then  the  angle  which 
the  required  line  makes  with  the  central  line  equals  one  of  the  base 
angles;  prove.] 

12.  To  construct  a  triangle,  being  given  the  vertical  angle,  one  of 
adjacent  sides,  and  the  perpendicular  from  the  vertex  to  the  base. 

[Show  that  the  foot  of  the  perpendicular  can  be  found  by  intersec- 
tion of  loci.] 

13.  K  two  circles  intersect,  then  any  common  tangent  subtends, 
at  the  common  points,  angles  which  are  supplemental. 

14.  If  a  common  tangent  is  drawn  to  two  circles,  and  if  each  point 
of  contact  is  joined  to  the  two  points  where  the  central  line  meets  the 
corresponding  circle,  then  the  two  chords  so  drawn  in  one  circle  are 
respectively  parallel  to  the  two  chords  in  the  other  circle. 

15.  Find  the  locus  of  a  point  such  that  the  tangent  from  it  to  a 
given  circle  shall  be  equal  to  the  line  joining  it  to  a  given  point. 

[This  is  a  limiting  case  of  the  radical  axis  of  two  circles  (179)  when 
the  radius  of  one  of  the  circles  diminishes  so  that  the  circle  reduces  to 
a  point.] 

16.  Find  a  point  such  that  the  tangent  from  it  to  a  given  circle 
shall  be  equal  to  each  of  the  lines  joining  it  to  two  given  points. 

17.  Find  the  locus  of  the  center  of  a  circle  which  passes  through  a 
given  point  and  cuts  a  given  circle  orthogonally. 

[In  188  let  one  of  the  circles  reduce  to  a  point.] 

18.  Describe  a  circle  through  a  given  point  so  as  to  cut  two  given 
circles  orthogonally. 

19.  Describe  a  circle  through  two  given  points  so  as  to  cut  a  given 
circle  orthogonally. 

20.  Given  the  vertical  angle  and  the  altitude  of  a  triangle,  prove 
that  the  surface  is  a  minimum  when  the  triangle  is  isosceles  (170). 

21.  Given  the  vertical  angle  and  altitude  of  a  triangle,  when  is  the 
base  a  minimum  ? 

22.  The  inscribed  regular  hexagon  is  equivalent  to  three  fourths 
of  the  circumscribed  one,  to  half  the  circumscribed  equilateral  triangle, 
and  to  double  the  inscribed  one,  in  the  same  circle. 


BOOK   IV.  — RATIO   AND  PROPORTION 

1.  That  relation  between  two  magnitudes  which  is  ex- 
pressed by  the  word  ratio  will  receive  a  precise  definition 
after  certain  preliminary  notions  are  explained. 

The  principles  will  be  made  sufficiently  general  to  apply 
to  any  geometric  magnitudes  for  which  appropriate  methods 
of  comparison  have  been  given,  such  as  two  line-segments, 
two  angles,  the  surfaces  of  two  polygons,  etc. 

It  will  not  be  necessary  to  restrict  our  thoughts  even  to 
geometric  magnitudes.  The  notion  of  ratio  is  applicable  to 
any  magnitudes  for  which  the  words  equivalent,  greater, 
less,  sum,  difference,  etc.,  have  a  definite  and  consistent 
meaning.  Such  magnitudes  are  found  in  the  sciences  that 
deal  with  number,  weight,  velocity,  probability,  etc. 

MULTIPLES  AND  MEASURES 

Definitions 

2.  Multiples.  If  any  number  of  equivalent  magnitudes 
are  added  together,  then  their  sum  is  called  a  multiple  of 
any  one  of  them. 

If  any  magnitude  P  is  equivalent  to  the  sum  of  n  magni- 
tudes each  equivalent  to  ^,  then  P  is  said  to  be  equivalent 
to  n  times  the  magnitude  A^  or  to  the  nth  multiple  of  A, 
which  is  sometimes  denoted  by  the  symbol  n-  A  ov  nA. 

Thus  the  double  of  A,  previously  defined,  means  the  same 
as  twice  A,  or  the  second  multiple  of  A.  We  may  regard  A 
itself  as  once  A,  and  call  it  the  first  multiple  of  A. 

3.  Series  of  multiples.  The  magnitudes  denoted  by  the 
symbols  ^^  2.1,  3^,  4^,  "-nA,  ... 

242 


MULTIPLES  AND  MEASURES  243 

may  be  thought  of  as  formed  by  beginning  with  the  magni- 
tude A  and  successively  adding  other  magnitudes  equivalent 
to  A,  as  often  as  desired.  The  whole  set  of  magnitudes  so 
thought  of  is  called  the  series  of  multiples  of  A. 

In  considering  the  mutual  relations  of  certain  magnitudes, 
their  series  of  mutiples  will  play  an  important  part. 

4.  Magnitudes  of  the  same  kind.  Two  magnitudes  will 
be  said  to  be  of  the  same  kind  when  their  two  series  of 
multiples  can  be  directly  compared  so  as  to  test  the  equiva- 
lence or  non-equivalence  of  any  of  them. 

In  particular  two  geometric  magnitudes  A  and  B  will  be 
said  to  be  of  the  same  kind  when  it  is  possible  to  compare 
the  multiples  of  A  with  the  multiples  of  B  by  means  of 
superposition.  Such,  for  instance,  are  two  line-segments, 
two  angles,  two  equiradial  arcs,  the  surfaces  of  two  polygons  ; 
but  not  a  straight  line  and  a  curved  line,  nor  two  arcs  of 
unequal  circles,  nor  the  surfaces  of  a  circle  and  a  polygon. 

For  a  simple  example  of  the  comparison  of  two  series  of 
multiples,  the  student  may  glance  forward  to  the  figure  in 
Art.  12,  in  which  the  successive  multiples  of  two  line-seg- 
ments A  and  B  are  laid  off  on  an  indefinite  line,  all  begin- 
ning at  the  same  point  0. 

Again,  the  natural  numbers  1,  2,  3,  4,  "-n,  ..., 

which  we  have  used  to  indicate  the  order  in  a  series  of  mul- 
tiples, belong  to  another  class  of  magnitudes  called  numeri- 
cal TYiainitudes.  The  terms  equivalent  (or  equal), 
greater,  less,  sum,  multiple,  etc.,  have  definite  meanings 
when  applied  to  them.  Thus,  the  number  3  has  its  series 
of  multiples,  3^  6,  9,  12,  ••., 

and  the  number  n  has  its  series  of  multiples, 

71,  2w,  3n,  •••,  pn,  •••. 

Any  two  natural  numbers  are  magnitudes  of  the  same 
kind,  since  they,  or  any  of  their  multiples,  are  directly 
comparable. 


244  PLANE  GEOMETRY  — BOOK  IV 

When  any  two  magnitudes  are  mentioned  together,  they 
are  understood  to  be  of  the  same  kind,  unless  otherwise 
stated.    Numerical  magnitudes  are  denoted  by  small  letters. 

5.  Common  multiple.  If  it  should  happen  that  some  mul- 
tiple of  A  is  also  a  multiple  of  5,  then  this  is  said  to  be  a 
common  multiple  of  A  and  B. 

It  rarely  happens  that  two  geometrical  magnitudes  taken 
at  random  have  a  common  multiple ;  but  if  they  have  one, 
then  they  have  an  indefinite  number  of  other  common  mul- 
tiples. For  instance,  if  the  magnitudes  A  and  B  have  a 
common  multiple  P  which  is  equivalent  to  2  A  and  also  to 
3  Bf  then  the  double  of  P  is  equivalent  to  4  A  and  also  to 
6b,  and  is  therefore  a  common  multiple  of  A  and  B. 

The  least  magnitude  (if  any)  which  is  a  common  multiple 
of  two  given  magnitudes  is  called  their  least  common 
multiple.  The  other  common  multiples  could  be  obtained 
by  starting  with  the  least  common  multiple,  and  then  form- 
ing its  series  of  multiples. 

In  the  figure  of  Art.  12,  if  ^  and  B  have  a  common  mul- 
tiple, this  fact  will  be  shown  by  the  coincidence  of  two  of 
the  points  of  division ;  and  such  coincidence  will  recur  at 
regular  intervals. 

6.  Submultiples  or  measures.  If  one  magnitude  is  a 
multiple  of  another,  then  the  latter  is  said 'to  be  a  sub- 
multiple  of  the  former. 

Thus  the  series  of  submultiples  of  a  magnitude  Aj  in 
descending  order  of  size,  are :  one  half  of  A,  one  third  of 
A,  one  fourth  of  ^,  . . .  one  nth  of  ^,  .  . . 

A  submultiple  of  A  is  often  called  a  measure  of  A,  be- 
cause it  is  contained  in  ^  a  certain  number  of  times  without 
remainder.  It  is  also  called  an  aliquot  part  of  A,  because 
it  may  be  obtained  by  dividing  A  into  a  certain  number  of 
equal  or  equivalent  parts.  We  can  use  the  term  that  seems 
most  suggestive  in  any  particular  connection.  The  magni- 
tude A  itself  may  be  included  among  the  measures  of  A. 


MULTIPLES  AND  MEASURES  245 

7.  Common  measure.  If  it  should  happen  that  some  meas- 
ure of  A  is  also  a  measure  of  B,  then  this  is  said  to  be  a 
comiTion  measure  of  A  and  B.  Two  magnitudes  that  have 
a  common  measure  are  said  to  be  commensuraJ)le. 

It  rarely  happens  that  two  geometrical  magnitudes  taken 
at  random  have  a  common  measure ;  but  if  they  have  one, 
then  they  have  an  indefinite  number  of  other  common 
measures.  For  instance,  if  the  magnitudes  A  and  B  have  a 
common  measure,  M,  which  is  contained  just  three  times  in 
A  and  five  times  in  B,  then  the  half  of  M  is  contained  just 
six  times  in  A  and  ten  times  in  B,  and  is  therefore  a  com- 
mon measure  of  A  and  B. 

The  greatest  magnitude  which  is  contained  a  whole  num- 
ber of  times  in  each  of  two  commensurable  magnitudes  is 
called  their  greatest  eoTmnon  measure.  The  other  com- 
mon measures  could  be  obtained  by  starting  with  the  greatest 
common  measure  and  then  taking  its  series  of  submultiples. 

8.  Like  multiples  and  like  measures.  If  there  are  any  two 
given  magnitudes  (not  necessarily  of  the  same  kind),  and  if 
two  multiples  of  them  are  formed  by  taking  each  of  them 
the  same  number  of  times,  then  the  two  resulting  magni- 
tudes are  called  liJce  multiples  of  the  two  given  magni- 
tudes, and  the  given  magnitudes  are  called  like  measures 
of  the  resulting  magnitudes. 

Thus,  if  the  magnitudes  A  and  X  are  each  taken  n  times, 
then  the  resulting  magnitudes  nA  and  nX  are  called  like 
multiples  of  A  and  X,  and  the  magnitudes  A  and  X  are 
called  like  measures  of  nA  and  nX. 

Properties  of  Multiples  and  Measures 

9.  The  following  general  statements  are  immediate  infer- 
ences from  the  preceding  definitions.  They  apply  to  any 
two  magnitudes  of  the  same  kind,  and  are  verified  by  direct 
comparison.  A  good  illustration  is  furnished  by  two  line- 
segments,  or  two  whole  numbers. 


246  PLANE  GEOMETRY  —  BOOK  IV 

1.  Two  magnitudes  of  the  same  kind  are  such  that  some 
multiple  of  one  is  greater  than  any  given  multiple  of  the  other. 

2.  According  as  one  magnitude  is  greater  than,  equivalent 
to,  or  less  than  another,  so  is  any  multiple  of  the  first  greater 
than,  equivalent  to,  or  less  than  the  like  multiple  of  the 
other. 

3.  According  as  one  magnitude  is  greater  than,  equivalent 
to,  or  less  than  another,  so  is  any  measure  of  the  first  greater 
than,  equivalent  to,  or  less  than  the  like  measure  of  the 
other. 

This  statement  is  converse  to  the  preceding;  they  can  be  put 
together  in  a  single  order-theorem  thus : 

The  two  pairs  of  magnitudes 

A,  B 
and  mA^  mB 

are  in  the  same  order  of  size. 

4.  Any  multiple  of  the  sum  of  two  or  more  magnitudes 
is  equivalent  to  the  sum  of  like  multiples  of  these  magnitudes. 

Symbolically :        m{A  -\-  B  ■\-C)<^mA  -\-  mB  -\-  mC. 

5.  Any  multiple  of  the  difference  of  two  magnitudes  is 
equivalent  to  the  difference  of  their  like  multiples. 

Symbolically  :        m{A  —  B)<>mA  —  mB. 

6.  A  common  measure  of  two  magnitudes  is  a  measure  of 
their  sum,  and  of  their  difference. 

7.  A  measure  of  any  magnitude  is  a  measure  of  any 
multiple  of  the  same  magnitude. 

8.  A  multiple  of  any  magnitude  is  a  multiple  of  any 
measure  of  the  same  magnitude. 

9.  The  mth  multiple  of  the  7ith  multiple  of  any  magnitude 
is  equivalent  to  the  mnth  multiple  of  the  same  magnitude. 

10.  The  mth  multiple  of  the  nth.  multiple  of  any  magni- 
tude is  equivalent  to  the  nth  multiple  of  the  mth  multiple  of 
the  same  magnitude. 

For  instance,  the  third  multiple  of  the  fourth  multiple  of  A  is 
equivalent  to  the  fourth  multiple  of  the  third  multiple  of  -4,  each  being 
equivalent  to  twelve  times  A. 


MULTIPLES   AND  MEASURES  247 

11.  According  as  m  is  greater  than,  equal  to,  or  less  than 
n,  so  is  the  mth  multiple  of  any  magnitude  greater  than, 
equivalent  to,  or  less  than  the  nth.  multiple  of  the  same 
magnitude;    and  conversely. 

These  two  converse  statements  may  be  expressed  as  an  order- 
theorem  thus : 

The  two  pairs  of  magnitudes 

w,  n 
and  mA,  nA 

are  in  the  same  order  of  size. 

EXAMPLES    FOR   ILLUSTRATION 

10.  The  following  examples  are  inserted  here  to  illustrate 
the  preceding  principles  ;  but  they  are  not  essential  to  the 
understanding  of  the  subsequent  articles. 

1.  If  the  mth  multiple  of  A  is  equivalent  to  the 
nth  multiple  of  B,  then  the  nth  suhmultiple  of  A 
is  equivalent  to  the  mth  suhmultiple  of  B. 

Denote  the  nth  suhmultiple  of  ^  by  ^',  and  the  mth  sub- 
multiple  of  Bhj  B\  then 

A^7iA',    B^mB'. 

Take  m  times  each  of  the  first  pair,  and  n  times  each  of 
the  second  pair,  then 

mA  =0=  mnA',  nB  ^  nmB'.  [9  (2,  9) 

Now,  by  hypothesis,  mA  is  equivalent  to  nB.     Hence. 

mnA '  =0=  nmB '  =c=  mnB '.  [9  (10) 

Take  the  mnth  suhmultiple  of  each  of  these  equivalents. 
Then  a'^b'.  [9(3) 

Cor.  I.  If  two  magnitudes  have  a  common  multiple,  then 
they  are  commensurable. 

Cor.  2.  If  two  magnitudes  are  incommensurable,  then  they 
have  no  common  multiple. 

Ex.  Any  two  whole  numbers,  m  and  w,  have  a  common  multiple 
mn,  and  a  common  measure  unity. 

«MCM.   ELEM.  GEOM. 17 


248  PLANE  GEOMETRY — BOOK  IV 

2.  If  the  nth  suhmultiple  of  A  is  equivalent  to  the 
mth  suhmultiple  of  B,  then  the  mth  multiple  of  A  is 
equivalent  to  the  nth  multiple  of  B, 

Outline.     With  the  same  notation  as  before 
A<>nA\     B^mB'; 
and  we  are  given  ^ '  =0=  5 ' ; 

to  prove  mA  =onB. 

Cor.  I.  If  two  magnitudes  are  commensurable,  then  they 
have  a  common  multiple. 

Cor.  2.  If  two  magnitudes  have  no  common  multiple,  then 
they  are  incommensurable. 

3.  To  find  whether  two  given  magnitudes  are  com^- 
mensura^le  or  not;  and,  if  so,  to  find  their  greatest 
common  measure. 

For  convenience,  let  the  magnitudes  be  represented  by  the 
lines  AE  and  FK. 

A  CDS 


F  G  H        K 

(a)  From  the  greater  AE  take  away  as  many  parts  as 
possible  each  equivalent  to  the  less  FK.  If  there  is  a 
remainder,  as  CE,  take  away  from  FK  as  many  parts  as 
possible  each  equivalent  to  CE.  If  there  is  a  second  re- 
mainder, as  HK,  take  away  from  the  preceding  remainder,  CE, 
as  many  parts  as  possible,  each  equivalent  to  HK,  and  so  on. 

It  is  evident  that  this  process  will  terminate  only  when 
a  remainder  is  obtained  which  is  a  measure  of  the  remainder 
immediately  preceding. 

(6)  If  this  process  terminates,  then  the  two  given  magnitudes 
are  commensurable,  and  the  last  remainder  is  their  greatest 
common  measure. 

First,  to  prove  that  the  last  remainder  is  a  common 
measure. 


MULTIPLES  AND  MEASURES  249 

Suppose  that  HK  is  the  last  remainder.  Then,  by  hypothe- 
sis, UK  is  a  measure  of  CE,  and  hence  of  FH,  which  is  a 
multiple  of  CE  [9  (7)] ;  therefore  HK  is  also  a  measure  of 
FK,  which  is  the  sum  of  FH  and  HK  [9-  (6)],  and  therefore 
HK  is  a  measure  of  ^c,  which  is  some  multiple  of  FK,  by 
construction ;  hence,  again,  HK  is  a  measure  of  AE,  which 
is  the  sum  of  ^C  and  GE-^  therefore  HK  is  a  common 
measure  of  FK  and  AE. 

Next,  to  prove  that  HK  is  the  greatest  common  measure 
of  FK  and  AE. 

Every  measure  of  Fir  is  a  measure  of  its  multiple  AC 
[9  (7)];  hence  every  common  measure  of  FK  and  AE  is  a 
common  measure  of  ^C  and  AE,  and  therefore  a  measure  of 
their  difference  CE,  and  therefore  of  FH,  which  is  a  multiple 
of  CE;  hence  every  common  measure  of  FK  and  AE  is  a 
common  measure  of  FK  and  FH,  and  therefore  a  measure  of 
their  difference  HK  [9  (6)]. 

Hence  no  common  measure  of  FK  and  AE  can  exceed  ifiT. 
Therefore  HK  is  the  greatest  common  measure  of  FK  and  ^F. 

(c)  i/*  ^^.e  ^?/;o  magnitudes  have  a  common  measure,  then  the 
process  in  (a)  will  terminate. 

Proof.  Any  common  measure  is  a  measure  of  each  re- 
mainder, as  shown  above.  Now  any  remainder  is  evidently 
less  than  half  the  second  preceding  remainder;  hence,  if 
the  process  does  not  terminate,  a  remainder  will  be  reached 
which  is  less  than  any  assigned  magnitude,  and  therefore 
less  than  the  greatest  common  measure ;  but  this  is  impos- 
sible, since  the  greatest  common  measure  is  a  measure  of 
every  remainder.  Therefore  the  process  does  terminate  if 
the  two  given  magnitudes  have  a  common  measure. 

(cT)  If  the  process  in  (a)  does  not  terminate,  then  the  two 
magnitudes  are  incommensurable. 

For  if  they  were  commensurable  the  process  would  ter- 
minate (c). 


250 


PLANE  GEOMETRY — BOOK  IV 


We  conclude  from  (b),  (c),  (d),  that  the  process  in  (a)  is  a 
complete  test  of  commensurability,  and  that  it  furnishes  the 
greatest  common  measure  when  one  exists. 

Note.  The  same  principles  can  be  used  in  finding  the  greatest 
common  measure  of  two  polygons ;  but  the  actual  process  may  be 
difficult  in  certain  cases.  In  the  case  of  two  numerical  magnitudes 
the  method  is  easily  applied,  and  corresponds  to  the  ordinary  arith- 
metical rule. 

4.  The  side  and  diagonal  of  a  square  are  incom- 
mensurable. 

Let  ABCDhQdi  square,  whose  side  is  AB,  and  diagonal  AC. 

To  prove  that  AB  and  AC  have 
no  common  measure.  /'^\^ 

On  AC  lay  off  a  part  AE  equal 
to  AB.  The  remainder,  EC^  is 
less  than  AB  (I.  88). 

Hence  AB  \%  contained  once  in 
AC  with  a  remainder  EC. 

Draw  EF  perpendicular  to  EC^ 
and  join  AF.  Since  AE  equals 
AD,  the  right  triangles  AEF  and 
ADF  are  equal  (I.  98).  Therefore 
EF  equals  FD. 

Again,  since  the  right  triangle  CEF  is  isosceles,  the  lines 
CE,  EFj  and  FD  are  equal. 

Lay  off  FG  equal  to  FD.  Then  EC  is  contained  twice  in 
the  side  DC,  with  remainder  GC. 

This  remainder  GC  is  for  a  similar  reason  contained  twice 
in  EC  with  remainder  EC;  and  so  on. 

Hence  this  process  of  finding  the  greatest  common  measure 
repeats  itself  indefinitely,  and  will  never  terminate. 

Therefore  AC  and  AB  are  incommensurable  [10  (3)]. 

Ex.    The  squares  on  ^C  and  AB  are  commensurable. 
Note.     Unless  two  magnitudes  are  specially  selected,  they  are  in  all 
probability  incommensurable.    Commensurability  is  a  rare  exception. 


MULTIPLES  AND  MEASURES  251 

Scale  op  Relation 

11.  Definition.  If  there  are  two  magnitudes,  A  and  B,  of 
the  same  kind,  and  if  their  two  series  of  multiples,  namely, 
A,  2 A,  Sa,  ..-,  and  B,  2b,  Sb,  •••,  are  supposed  written  in 
a  single  scale  in  the  ascending  order  of  size  of  all  these  mul- 
tiples, the  resulting  arrangement  is  called  the  scale  of 
relation  of  the  two  magnitudes  A  and  B. 

12.  Two  line-segments.  The  scale  of  relation  of  two 
given  line-segments  is  easy  to  construct,  and  furnishes  a 
good  illustration  of  the  definition. 

Let  the  two  given  lines  be  A  and  B, 

B 


B  SB  S^  fB  SB  6B 

2^A         S         A        Ya         6  a         7a        <?k       "^ 


Take  an  indefinite  line  OX,  and,  beginning  at  0,  lay  off 
successive  segments  each  equal  to  A.  Mark  the  consecu- 
tive points  of  division  with  the  symbols  A,  2 A,  SA,  «... 
Beginning  again  at  o,  lay  off  successive  segments  each 
equal  to  B ;  and  mark  the  points  of  division  with  the  sym- 
bols B,  2  b,  3  b,  ..-. 

The  order  in  which  the  symbols  occur  on  the  line  evi- 
dently shows  the  order  of  succession  of  the  various  multiples 
of  the  two  given  magnitudes,  all  arranged  in  one  scale  in 
the  ascending  order  of  their  size.  This  order  of  succession 
constitutes  the  scale  of  relation  of  the  two  given  magnitudes 
A  and  B.  One  use  of  such  a  scale  is  to  tell  between  what 
two  consecutive  multiples  of  B  any  assigned  multiple  of  A 
lies.  For  instance,  in  the  figure  8  ^  is  greater  than  5  B  and 
less  than  6b.  The  scale  will  also  show  whether  A  and  B 
have  any  common  multiple.  If  two  of  the  points  of  divi- 
sion happen  to  coincide  at  any  point  P,  then  the  line  OP  is 
a  common  multiple  of  the  two  given  lines.     If  A  and  B  are 


252  PLANE  GEOMETRY  —  BOOK  IV 

incommensurable,  they  have  no  common  multiple,  and  none 
of  the  points  of  division  will  coincide. 

13.  Two  numerical  magnitudes.  As  another  illustration, 
the  scale  of  relation  of  two  given  whole  numbers  may  be 
found  by  a  method  similar  to  that  just  given.  For  instance, 
the  scale  of  relation  of  the  numbers  3  and  14  may  be  ar- 
ranged as  follows  (letting  t  stand  for  3  and  /  for  14) : 

t,  2t,  3t,  4:t,f,  5t,  6t,  7t,  St,  9t,  2f,  lOi,  lit,  12t,  ISt, 
14«  =  3/,  15 ty  16 ty  17 1,  ISt,  4/,  .... 

RATIO 
On  the  Notion  of  Ratio 

14.  Definitions.  That  relation  of  two  magnitudes  of  the 
same  kind  which  is  exhibited  by  the  order  of  succession  of 
their  multiples  when  arranged  in  one  ascending  scale  is 
called  the  ratio  of  one  magnitude  to  the  other. 

The  ratio  of  a  magnitude  A  to  another  magnitude  B  of 
the  same  kind  is  denoted  by  the  symbol  A  :  B ;  and  A  is 
called  the  antecedent  and  B  the  consequent  of  the  ratio. 

If  there  are  two  other  magnitudes  of  the  same  kind,  X 
and  r  (not  necessarily  of  the  same  kind  as  A  and  B),  the 
definitions  of  the  next  two  articles  furnish  a  basis  for  a 
comparison  of  the  two  ratios  A  :  B  and  X :  T,  so  that  the 
terms  equal,  greater,  and  less  may  have  definite  meanings 
when  applied  to  any  two  ratios. 

The  two  antecedents  are  said  to  be  homologous  terms  in 
the  two  ratios,  and  so  are  the  two  consequents. 

15.  Equal  ratios.  The  two  ratios  A  :  B  and  X  :  Y  are  said 
to  be  equal  ratios,  if,  on  comparison  of  the  scale  of  rela- 
tion of  A  and  B  with  the  scale  of  relation  of  X  and  r,  it  is 
found  that  the  successive  multiples  of  A  and  B  interlie  in 
the  same  order  in  the  first  scale  as  the  corresponding  multi- 
ples of  X  and  r  do  in  the  second  scale. 


BATIO  253 

In  the  same  case  the  two  scales  of  relation  are  said  to  be 
similar  to  each  other. 

In  applying  this  definition,  the  similarity  of  the  scale  of 
A  and  B  with  the  scale  of  X  and  Y  can  be  established  in  the 
following  manner : 

Take  any  like  multiples  of  the  antecedents,  say 
mA,  mX\ 
and  take  any  like  multiples  of  the  consequents,  say 

nB,  nY\ 
then,  to  prove  that  the  two  scales  are  similar  throughout, 
we  have  to  show  that  the  two  pairs  of  multiples 

mA,  nB 
and  mX,  nY 

are  in  the  same  order  of  size,  no  matter  what  whole  numbers 
m  and  n  are. 

The  phrase  "  in  the  same  order  of  size  "  has  been  defined 
and  fully  illustrated  in  connection  with  "order-theorems" 
in  III.  50.  The  student  is  requested  now  to  review  that 
article,  and  then  to  read  the  simple  application  of  the  above 
definition  of  equal  ratios,  which  will  be  found  in  art.  90, 
Book  V,  where  it  is  proved  that 

If  any  two  rectangles  have  equal  altitudes,  then 
the  ratio  of  the  rectangles  is  equal  to  the  ratio  of 
their  bases. 

The  method  may  be  stated  in  outline  as  follows : 

Let  the  rectangles  be  denoted  by  R  and  R\  and  their  bases 
by  h  and  h\  then  we  have  to  prove  that  the  ratio  R:  R'  is 
equal  to  the  ratio  6:6'. 

Of  the  base  6  we  take  any  multiple  mh ;  and  then  show 
that  the  rectangle  standing  on  the  base  mh,  and  having  the 
given  altitude,  is  equivalent  to  mR. 

Of  the  base  6'  we  take  any  multiple  n6';  and  then  show 
that  the  rectangle  standing  on  the  base  w6',  and  having  the 
given  altitude,  is  equivalent  to  7iR'. 


254  PLANE  GEOMETRY — BOOK  IV 

We  then  quote  the  order-theorem  that  any  two  rectangles 
of  the  same  altitude  are  in  the  same  order  of  size  as  their 
bases  are ;  and  thence  infer  that  the  two  pairs  of  multiples 

mR,  nR' 
and  mbj  nb' 

are  in  the  same  order  of  size,  whatever  whole  numbers  m 
and  n  may  be. 

We  conclude  that  the  scale  of  relation  of  R  and  R'  is 
similar  throughoi^t  its  whole  extent  to  the  scale  of  relation 
of  b  and  b' ;  and  hence,  by  definitions  of  equal  ratios,  that 

R:R'==b:b'. 

16.  Unequal  ratios.  The  ratio  A:  B  is  said  to  be  greater 
than  the  ratio  X :  r,  if,  on  comparison  of  the  scale  of  rela- 
tion of  A  and  B  with  the  scale  of  relation  of  X  and  Y,  it  is 
found  that  some  multiple  of  A  has  a  more  advanced  position 
among  the  multiples  of  B  than  the  like  multiple  of  X  has 
among  the  multiples  of  Y. 

In  other  words,  the  ratio  A:  B  is  said  to  be  greater  than 
the  ratio  X  :  Y  when  it  is  possible  to  find  any  single  pair  of 
whole  numbers  m  and  n  such  that 

mA  is  greater  than  n5,  and 
mx  is  not  greater  than  wF, 

or  else  such  that 

mA  is  equivalent  to  nB,  and 
mX  is  less  than  nY. 

In  the  same  case  the  ratio  X :  F  is  said  to  be  less  than 
the  ratio  A  :  B.     This  is  expressed  symbolically  by 

A:B>X:Y    or    X  :  Y  <  A  :  B. 

Note.  A  ratio  is  not  to  be  confused  with  its  number-correspondent, 
introduced  later,  in  mensuration  (VI.  14).  It  will  appear  that  the 
general  theory  of  ratio  furnishes  a  natural  and  logical  basis  for  the 
science  of  numerical  measurement,  which  is  one  of  its  applications. 


RATIO  255 

Properties  of  Ratios 

17.  In  the  following  articles,  magnitudes  of  the  same  kind 
will  usually  be  denoted  by  adjacent  letters  of  the  alphabet ; 
and  magnitudes  that  are  not  necessarily  of  the  same  kind 
will  be  denoted  by  non-adjacent  letters.  For  instance,  if 
there  are  three  ratios  whose  terms  are  not  restricted  to  be 
of  the  same  kind,  they  may  be  represented  by  A  :  B,  P  :  Q, 
X :  Y.     A  small  letter  will  denote  a  whole  number. 

PRINCIPLES    OF    EQUALITY    AND    INEQUALITY 

18.  It  has  been  seen  that  a  ratio  is  not  a  magnitude,  but  a 
relation  between  two  magnitudes.*  Two  ratios  can,  however, 
be  compared  with  each  other  by  means  of  the  conven- 
tions laid  down  in  15,  16;  and  it  will  be  proved  in  theo- 
rems 1  and  2  that  the  definitions  of  the  words  equal, 
greater,  and  less,  as  applied  to  ratios,  lead  to  principles 
that  correspond  to  the  axioms  of  equality  and  inequality. 

Principle  of  equality, 

19.  Theorem  1.  If  two  ratios  are  each  equal  to  the 
same  third  ratio,  then  they  are  equal  to  each  other. 

Let  A  :  B  =  X  :  Yj 

and  P'.Q  =  X'.Y'^ 

to  prove  A  :  B  =  P '.  Q. 

The  multiples  of  A  and  B  have  the  same  inter-order  as 
the  multiples  of  X  and  F;  and  the  same  thing  is  true  of  the 
multiples  of  P  and  Q ;  therefore,  the  multiples  of  A  and  B 
have  the  same  inter-order  as  those  of  P  and  Q;  thus  the 
scales  of  relation  are  similar ;  hence,  by  definition  (15), 

A'.B  =  P\Q. 

*  This  statement  applies  even  to  numerical  ratios  ;  the  ratio  m  :  n  is 

distinct  from  the  quotient  — ,   which  is  its    number-correspondent 

n 
(VI.  14). 


256  PLANE  GEOMETRY  —  BOOK  IV 

Principle  of  inequality, 

20.  Theorem  2.  If  two  ratios  are  equal,  then  any 
third  ratio  which  is  less  than  one  of  them  is  also  less 
than  the  other. 

Let  A.  B  =  X:Yj 

and  P'.Q<X:Y', 

to  prove  P  :  Q  <A:  B. 

Since  P:Q<X:Yy 

hence,  by  the  definition  of  unequal  ratios,  some  multiple  of 
P,  say  the  ?7ith,  occupies  a  less  advanced  position  among  the 
multiples  of  Q  than  the  mth  multiple  of  X  does  among  the 
multiples  of  F;  therefore,  by  the  first  part  of  the  hypothe- 
sis, the  mth  multiple  of  P  occupies  a  less  advanced  position 
among  the  multiples  of  Q  than  the  mth  multiple  of  A  does 
among  the  multiples  of  B.  Hence,  by  definition  (16), 
P:Q<A:B. 

21.  Cor.  If  two  ratios  are  equal,  then  any  third  ratio 
which  is  greater  than  one  is  also  greater  than  the  other. 

RECIPROCAL    RATIOS 

22.  Definition.  Two  ratios  are  said  to  be  reciprocal  to 
each  other  when  the  antecedent  of  each  ratio  is  the  conse- 
quent of  the  other. 

Thus,  the  ratios  A  :  B  and  B  :  A  are  reciprocal  to  each 
other. 

Reciprocals  of  equal  ratios, 

23.  Theorem  3.  //  two  ratios  are  equal,  then  their 
reciprocal  ratios  are  equal. 

Given  A\B  =  X'.  F; 

to  prove  B  .  A  =  Y :  X. 

From  the  hypothesis,  any  multiple  of  A  occupies  a  posi- 
tion among  the  multiples  of  B,  similar  to  that  which  the  like 
multiple  of  X  occupies  among  the  multiples  of  Y. 


RATIO  257 

Hence  any  multiple  of  B  occupies  a  position  among  the 
multiples  of  A  similar  to  that  which  the  like  multiple  of  Y 
occupies  among  the  multiples  of  X. 

Therefore  the  ratio  ^  :  ^  is  equal  to  the  ratio  Y :  X  (15). 

Reciprocals  of  unequal  ratios, 

24.  Theorem  4.  If  one  ratio  is  greater  than  an- 
other, tJwn  the  reciprocal  of  the  first  is  less  than 
the  reciprocal  of  the  second. 

Given  A:  B>  X:  F; 

to  prove  B  :  A  <  Y '.  X. 

From  the  hypothesis,  some  multiple  of  A  occupies  a  more 
advanced  position  among  the  multiples  of  B  than  the  like 
multiple  of  X  does  among  those  of  Y. 

Hence  some  multiple  of  B  occupies  a  less  advanced  posi- 
tion among  the  multiples  of  A  than  the  like  multiple  of  Y 
does  among  the  multiples  of  X. 

Therefore,  by  definition,  the  ratio  B  :  A  i^  less  than  the 
ratio  r  :  X  (16). 

EQUIVALENCE    OF    ANTECEDENTS    OR    CONSEQUENTS 

This  group  of  theorems  is  concerned  with  the  comparison 
of  ratios  whose  antecedents  or  consequents  are  equivalent. 

Equivalent  antecedents  and  consequents, 

25.  Theorem  5.  If  two  ratios  have  equivalent  ante- 
cedents and  equivalent  consequents,  then  the  ratios 
are  equal. 

Griven  A  equivalent  to  A\  and  B  equivalent  to  5'; 
to  prove  A  :  B  =  A' :  b'. 

Since  the  multiples  of  A  are  equivalent  to  the  like  mul- 
tiples of  ^'  [9  (2)]  ;  and  the  multiples  of  B  are  equivalent 
to  the  like  multiples  oi  B';  therefore,  the  multiples  of  A  are 
distributed  among  the  multiples  of  B  in  the  same  inter-order 
as  the  multiples  of  A'  are  among  those  of  B'. 

Hence, by  definition,  the  ratio  A  :  B  equals  the  ratio  A^ :  bK 


258  PLANE  GEOMETRY — BOOK  IV 

26.  According  to  the  theorem  just  proved,  a  ratio  is  not 
altered  when  either  its  antecedent  or  its  consequent  is  re- 
placed by  an  equivalent  magnitude.  For  this  reason  two 
equivalent  magnitudes  will  sometimes  be  denoted  by  the 
same  letter;  and  the  symbol  of  equality  (=)  will  be  used 
as  the  symbol  of  equivalence.  The  symbols  for  greater 
than  (>)  and  for  less  than  (<)  will  also  be  used. 

Equivalent  consequents,  unequivalent  antecedents, 

27.  Theorem  6.  If  two  ratios  have  equivalent 
consequents,  then  the  one  that  has  the  greater  ante- 
cedent is  the  grea;ter  ratio. 

Let  each  of  the  consequents  be  equivalent  to  a  magnitude 
C;  and  let  the  antecedents  be  A  and  B. 

Given  A>B', 

to  prove  A:C>  B  :  C. 

Let  A  exceed  5  by  a  magnitude  .Y. 

Take  a  multiple  of  X  that  shall  exceed  c ;  and  let  mX  be 
such  a  multiple.  Take  the  like  multiples  of  A  and  B.  Then 
mA  exceeds  mB  by  mX  [9  (5)]. 

Therefore  mA  exceeds  mB  by  more  than  C;  or,  in  other 
words,  mB  falls  short  of  mA  by  more  than  C. 

Next,  take  the  successive  multiples  of  C,  and  let  nC  be 
the  first  one  that  does  not  fall  short  of  mA. 

Then  either  mA  is  equivalent  to  nC,  or  else  mA  lies  be- 
tween nC  and  the  next  lower  multiple  of  C. 

But  mB  falls  short  of  mA  by  more  than  C ;  therefore  mB 
lies  in  a  less  advanced  interval  among  the  multiples  of  C 
than  mA  does. 

Hence  the  ratio  ^  :  C  is  greater  than  the  ratio  B :  C  (16). 

Combined  statement. 

28.  Cor.  I.  If  tivo  ratios  have  equivalent  consequents,  then 
according  as  the  first  antecedent  is  greater  than,  equivalent  to, 
or  less  than  the  second,  so  is  the  first  ratio  greater  than,  equal 
to,  or  less  than  the  second.     (Combination  of  25  and  27.) 


BATIO  259 

Converse  statement. 

29.  Cor.  2.  If  two  ratios  have  equivalent  consequents,  then 
according  as  the  first  ratio  is  greater  than,  equal  to,  or  less  than 
the  second,  so  is  the  first  antecedeiit  greater  than,  equivalent  to, 
or  less  than  the  second. 

Equivalent  antecedents,  unequivalent  consequents, 

30.  Theorem  7.  If  two  ratios  have  equivalent  ante- 
cedents, then  the  one  that  has  the  greater  conse- 
quent is  the  less  ratio. 

Let  each  of  the  antecedents  be  equivalent  to  A ;  and  let 
the  consequents  be  B  and  C. 

Given  ^  >  C ; 

to  prove  A:  B  <  A:  C. 

Take  B  and  G  as  antecedents,  and  compare  them  with  the 
same  consequent  A. 

Then,  since  B  is  greater  than  C, 
hence,  B  :  A>  C:  A,  [27 

Therefore,  by  taking  reciprocals, 

A'.B<A'.C.  [24 

Conibined  statement. 

31.  Cor.  I.  If  two  ratios  have  equivalent  antecedents,  then 
according  as  the  first  consequent  is  greater  than,  equivalent  to, 
or  less  than  the  second,  so  is  the  first  ratio  less  thari,  equal  to, 
or  greater  than  the  second.     (Combination  of  25  and  30.) 

Converse  statement. 

32.  Cor.  2.  If  two  ratios  have  equivalent  antecedents,  then 
according  as  the  first  ratio  is  less  than,  equal  to,  or  greater 
than  the  second,  so  is  the  first  consequent  greater  than,  equlvor- 
lent  to,  or  less  thari  the  second. 

Note.  It  follows  from  27  and  30  that  the  scale  of  relation  of  two 
given  magnitudes  is  altered  somewhere,  if  the  slightest  change  is  made 
in  either  one  of  the  given  magnitudes. 


260  PLANE  GEOMETRY  —  BOOK  IV 

Homologous  terms  of  equal  ratios  compared, 

33.  Theorem  8.  If  there  are  two  equal  ratios^  the 
four  magnitudes  being  of  the  same  kind,  then  a/i- 
cording  as  the  antecedent  of  tlxe  first  is  greater  than, 
equivalent  to,  or  less  than  the  antecedent  of  the  sec- 
ond, so  is  the  consequent  of  the  first  greater  than, 
equivalent  to,  or  less  than  the  consequent  of  the 
second. 

Let  Aj  BjCfDyhe  four  magnitudes  of  the  same  kmd,"such 
*l^t  A  '.  B  =  C :  D, 

To  prove  that  the  two  pairs 


^, 

C 

and 

B, 

D 

are  in 

the 

same 

.order  of 

size. 

First,  let  A  be  equivalent  to  C. 

Then  the  two  equal  ratios  written  above  have  equivalent 
antecedents ;  hence  their  consequents  are  equivalent, 
i.e.  B  is  equivalent  to  D.  [32 

Next,  let  A  be  greater  than  C. 

Take  A  and  c  as  antecedents,  and  compare  them  with  the 
same  consequent  5, 

then  A'.B>C.B^  [27 

but  A:B  =  C:Df  [hyp. 

therefore,  by  the  principle  of  inequality, 

C:  D>  C:B.  [20 

Since   these   unequal   ratios  have  the  same  antecedent, 
hence  the  less  ratio  has  the  greater  consequent, 
i.e,  B  is  greater  than  D.  [32 

Lastly,  let  A  be  less  than  c. 

The  student  may  prove  in  a  similar  way  that  in  this  case  B  is  less 
than  D. 

Thus  the  two  pairs  above  are  always  in  like  order  of  size. 


JRATIO  261 

33  (a).  Ex.  If  there  are  two  equal  ratios  A  :  B  and  X :  F,  then 
the  two  pairs 

A,  B 
and  X,  T 

are  in  the  same  order  of  size. 

(Put  m  =  n  =  1  in  the  order-statement  given  in  Article  15.) 

N.B.  This  principle,  which  follows  immediately  from  the  defini- 
tion of  equal  ratios,  is  not  to  be  confused  with  the  principle  in  33. 

RATIOS    OF    MULTIPLES 

34.  The  next  three  theorems  with  their  corollaries  relate 
to  the  comparison  of  multiples  of  given  magnitudes. 

LiTce  multiples  of  two  magnitudes, 

35.  Theorem  9.  The  ratio  of  like  multiples  of  two 
magnitudes  is  equal  to  the  ratio  of  the  magnitudes. 

Let  A  and  B  be  tv^^o  magnitudes  of  the  same  kind,  and  let 
pA  and  pB  be  any  like  multiples  of  A  and  B. 
To  prove  pA  :  pB  =  A  :  B. 

Take  the  mth  multiple  of  each  of  these  antecedents,  and 
the  nth.  multiple  of  each  consequent,  and  compare  the  order 
of  size  of  the  resulting  pairs  of  multiples, 

m  'pA,  n  'pB 
and  m  '  A,  n-  B. 

Since  the  mth  multiple  of  the  pth  multiple  is  equivalent 
to  the  pth  multiple  of  the  mth  multiple,  the  first  pair  may 
be  written  in  the  form 

P'mA,p'nB.  [9(10) 

Now  the  members  of  this  pair  are  like  multiples  of  mA 
and  nB ;  hence  this  pair  is  in  the  same  order  of  size  as  the 
pair  mA,  nB,  [9  (3) 

which  is  the  second  pair  of  multiples  above. 

Hence  the  above  pairs  of  multiples  are  in  the  same  order 
of  size  whatever  m  and  n  are. 

Therefore  pA:pB  =  A:B. 


262  PLANE  GEOMETRY — BOOK  IV 

Special  case. 

36.  Cor.  I.  The  ratio  of  two  whole  numbers  is  not  altered 
by  multiplying  both  of  them  by  the  same  number. 

37.  Cor.  2.     If  A:B  =  X:  Y, 

then    pA  :  pB  =  qX  :  qY, 

Multiples  of  one  magnitude. 

38.  Theorem  10.  The  ratio  of  the  mth  multiple  of 
any  magnitude  to  the  nth  multiple  of  the  same  mag- 
nitude is  equal  to  the  ratio  of  the  number  m  to  the 
number  n. 

To  prove  mA  :  nA  =  m  :n. 

Take  p  times  each  antecedent,  and  q  times  each  conse- 
quent, and  compare  the  order  of  size  of  the  two  pairs  of 

multiples 

p  •  mA,  q  •  nA 

and  p  -m,  q-n. 

The  first  pair  may  be  written  in  the  form 

pm  •  A,  qn  •  A,  [9  (9) 

and  these  two  multiples  of  A  are  in  the  same  order  of  size 
as  the  pair  of  numbers 

pm,  qn,  [9  (11) 

which  is  the  second  pair  of  multiples  above. 

Hence  the  two  pairs  of  multiples  above  are  in  the  same 
order  of  size  whatever  the  whole  numbers  p  and  q  are. 

Therefore,  by  the  definition  of  equal  ratios, 

mA  :nA  —  m:n. 

Another  statement  of  38. 

39.  Cor.  If  two  magnitudes  have  a  common  measure 
which  is  contained  m  times  in  the  first  magnitude  and  n  times 
in  the  second,  then  the  ratio  of  the  tivo  magnitudes  is  equal  to 
the  ratio  of  the  number  m  to  the  number  n. 


RATIO  263 

Equivalent  multiples, 

40.  Theorem  11.  If  the  ratio  of  one  magnitude  to 
another  is  equal  to  the  ratio  of  the  number  m  to  the 
number  n,  then  the  nth  multiple  of  the  first  magni- 
tude is  equivalent  to  the  mth  multiple  of  the  second. 

Given  A-.  B  =  m-.n] 

to  prove  nA  —mB. 

Take  the  nth  multiple  of  each  antecedent  and  the  mth 
multiple  of  each  consequent ;  then,  from  the  hypothesis,  the 
pairs  of  multiples 

n-  A,  m  •  B 

and  n-m,  m-n 

are  in  the  same  order  of  size ;  but  the  members  of  the  latter 
pair  are  equivalent;  therefore  the  members  of  the  former 
pair  are  equivalent. 

Hence  nA  =  mB. 

Uneqiiivalent  tnultiples, 

41.  Cor.  I.  If  the  ratio  of  tivo  magnitudes  is  greater  than 
the  ratio  of  two  ivhole  numbers  m  and  n,  then  the  7ith  multiple 
of  the  first  magnitude  is  greater  than  the  mth  multiple  of  the 
second. 

[Show  that  the  first  pair  of  multiples  above  are  then  in  descending 
order  (16).] 

Combined  statement. 

42.  Cor.  2.     According  as  A  :  B  >  =  <^m:n, 

so  is  nJ^  >  =  <  mB. 

Converse  statement. 

43.  Cor.  3.     According  as      nA  >  =  <  mB, 

so  is  A  :  B  '^  =  <m:n. 

MCM.    ELEM.    GEOM.  — 18 


264  PLANE  GEOMETRY  —  BOOK  IV 

PROPORTION 
Properties  of  a  Proportion 

44.  Definition.  A  proportion  is  a  statement  of  the 
equality  of  two  ratios,  2iS  A:  B  =  X:  Y. 

These  four  magnitudes  are  said  to  form  a  proportion,  of 
which  A  and  Y  are  the  extrejnes,  and  B  and  X  the  means ; 
and  Y  is  called  the  fourth  proportional  to  the  three  terms 
A  J  B,  and  X.  The  proportion  is  sometimes  read  thus :  A  is 
to,  i?  as  X  is  to  Y. 

The  next  three  theorems  are  concerned  with  the  establish- 
ment of  certain  general  "  rules  of  inference,"  by  which,  from 
a  given  proportion,  certain  other  proportions  can  be  at  once 
derived.  They  are  the  Rules  of  Equi-multiplication,  Alter- 
nation, and  Composition. 

Equi-tnultiplea  of  homologous  terms, 

45.  Theorem  12.  //  two  ratios  are  equal,  and  if 
any  like  multiples  of  the  antecedents  are  taken, 
and  also  any  like  multiples  of  the  consequents,  then 
the  m^ultiple  of  the  first  antecedent  is  to  tJie  mul- 
tiple of  the  first  consequ^ent  as  the  multiple  of  the 
second  antecedent  is  to  the  multiple  of  the  second 
consequent. 

Given  A-.B  —  X-.T-, 

to  prove  mA  :  nB  =  mx :  nY. 

To  compare  the  latter  two  ratios,  take  the  p\h.  multiple  of 
each  antecedent,  and  the  gth  multiple  of  each  consequent, 
and  compare  the  order  of  size  of  the  two  pairs  of  resulting 

multiples 

p  •  mA,  q  •  nB 
and  p'lnX,    q-nY. 

According  to  9  (9),  these  may  be  written  in  the  form 
pm  'A,  qri'  B 
and  ptn  *Xj  qn-  Y. 


PROPORTION  265 

Now  these  two  pairs  of  multiples  are  in  the  same  order  of 
size,  because  the  ratios  A  :  B  and  X  :  Y  are  equal. 

Therefore  the  former  pairs  of  multiples  are  in  the  same 
order  of  size,  whatever  whole  numbers  p  and  q  may  be. 

Hence  mA  :  itB  =  mX  :  nY. 

Special  case. 

46.  Cor.     Given         A  :  B  =  x  :  Y, 

and  mA  =  nB ; 

then  mX  =  nY. 

Note.    This  corollary  may  be  stated  in  words  as  follows : 

If  four  magnitudes  form  a  proportion ,  and  if  the  first  is  any 
multiple,  or  part,  or  multiple  of  a  part,  of  the  second,  then  the 
third  is  the  like  multiple,  or  part,  or  multiple  of  a  part,  of  the 
fourth. 

Rule  of  alternation. 

47.  Theorem  13.  If  four  magnitudes  of  the  same 
kind  form  a  proportion,  then  the  first  is  to  the  third 
as  the  second  is  to  the  fourth. 

Let  A,  B,  C,  D  be  four  magnitudes  of  the  same  kind  such 
*^^^  A:  B  =  C  :  D. 

To  prove  A:C  =  B:D. 

Since  the  ratio  of  two  magnitudes  equals  the  ratio  of  their 
like  multiples, 
hence  mA  :  mB  =  nC  :  nD.  [35 

Therefore,  by  comparison  of  homologous  terms  in  equal 
ratios,  the  two  pairs 

mA,  nC 
and  mB,  nD 

are  in  the  same  order  of  size.  [33 

Now  m  and  n  are  any  whole  numbers ;  hence,  by  definition 
of  equal  ratios,  A-  C  =  B  ■  D 


266  PLANK  GEOMETRY — BOOK  IV 

Rule  of  composition, 

48.  Theorem  14.  If  four  magnitudes  form  a  pro- 
portion, then  the  sum  of  the  first  and  second  is  to 
the  second  as  the  sujjv  of  the  third  and  fourth  is  to 
the  fourth. 

Given  A:B  =  X:Y; 

to  prove  A  +  B  :B  =  X-\-Y:Y. 

In  order  to  compare  the  latter  two  ratios  take  any  like 
multiples  of  the  antecedents,  and  any  like  multiples  of  the 
consequents ;  and  then  compare  the  order  of  size  in  the  two 
resulting  pairs 

m  (J  +  B),  nB 
and  m{x  -\-  r),  nY, 

First,  let  n  be  greater  than  m. 

The  order  of  the  first  pair  of  multiples  is  not  altered  by 
subtracting  inB  from  each ;  and  the  order  of  the  second  pair 
is  not  altered  by  subtracting  mY  from  each. 

Therefore  the  above  pairs  of  multiples  are  in  the  same 
order,  respectively,  as  the  pairs 

mA,  (71  —  m)B 
and  mA",  (n  —  m)  Y. 

Now,  from  the  hypothesis,  these  are  in  the  same  order  of 
size ;  hence  the  above  pairs  are  in  the  same  order  of  size. 

Next,  let  11  be  not  greater  than  m. 

Then  the  pairs  of  multiples  in  question  are  evidently  both 
in  descending  order. 

Therefore  m(A  -{-  B),  nB 

and  m(.Y-f  r),  nY 

are  always  in  the  same  order  of  size  whatever  m  and  n  are. 

Hence  A  -\-  B  :  B  =  X  +  Y  :Y. 

Rule  of  separation. 

49.  Cor.     In  the  same  case  A  —  B  :  B  =  X  —  Y:Y. 
The  complete  statement  and  proof  are  left  to  the  student. 


PBOPOBTION  267 

50.  It  is  convenient  to  insert  here  the  following  restate- 
ment of  theorem  3,  to  be  called  the  "  rule  of  reciprocation." 

51.  If  four  magnitudes  form  a  proportion,  then  the  second 
is  to  the  first  as  the  fourth  is  to  the  third. 

Two  OR  More  Proportions 

52.  The  next  three  theorems  are  concerned  with  rules  of 
inference  from  two  or  more  proportions.  They  are  the 
Rules  of  Combination,  of  Succession,  and  of  Addition. 

Rule  of  combination, 

53.  Theorem  15.  If  there  are  any  number  of  equal 
ratios,  all  the  magnitudes  being  of  the  same  hind, 
then  as  any  of  the  antecedents  is  to  its  consequent  so 
is  the  sum  of  all  the  antecedents  to  the  sum  of  all 
the  consequents. 

Given        A  :  B  =  A'  :B'  =  A"  :  B"-, 
to  prove         A  :  B  =  A  +  A'  -\-  a"  :  B  -\-  B'  -\-  B". 

From  the  hypothesis,  and  the  definition  of  equal  ratios, 
the  three  pairs  of  multiples 

mA,     nB 
mA',    nB' 
mA",  7iB" 
are  in  the  same  order  of  size ;  hence  the  pair 

m(A  -\-A'-\-  A"),  n(B  -{-  B'  +  B") 
is  also  in  the  same  order  as  any  of  the  preceding  pairs, 
whatever  m  and  n  are  (axioms  I.  25,  32 ;  II.  9,  10 ;  III.  38). 
Therefore       A  :  B  =  A  +  A'  +  A"  :  B  -\-  b'  -\-  B". 

54.  Definitions.  A  set  of  ratios  will  be  called  successive 
when  the  consequent  of  each  is  the  antecedent  of  the  next. 

The  first  antecedent  and  the  last  consequent  are  called  the 
extremes  of  the  set.  E.g.,  the  ratios  A  :  B,  B  :  C,  C  :  D,  D  :  E 
are  a  set  of  successive  ratios,  whose  extremes  are  A  and  E. 


268  PLANE  GEOMETRY  —  BOOK  IV 

Rule  of  succession, 

55.  Theorem  16.  //  there  are  any  nuinber  of  like 
magnitudes  and  an  equal  number  of  any  other  like 
magnitudes,  such  that  the  successive  ratios  in  tJie 
first  set  are  equivalent  respectively  to  the  correspond- 
ing successive  ratios  in  the  second  set,  then  the  ratios 
of  the  extremes  in  the  two  sets  are  equal. 

1.   Let  there  be  three  magnitudes  in  eaxjh  setj  and  let 

them  be 

Ay  B,  c 
and  X,  r,  z. 

Let  the  successive  ratios  A  :  B  and  5  :  c  be  equal  to  the 
successive  ratios  A' :  Y  and  Y :  Z,  respectively. 

To  prove  that  the  ratios  of  the  extremes  are  equal, 

i.e.  A',  c  =  X  \  z. 

Take  any  like  multiples  of  the  antecedents,  and  any  like 
multiples  of  the  consequents  ;  and  compare  the  order  of  size 
in  the  two  pairs 

mA,  nC 
and  mXy  nZ. 

First,  suppose  the  first  pair  to  be  in  descending  order, 

i.e.  mA  >  nC ; 

then,  comparing  each  of  these  magnitudes  with  the  same 
consequent  mB,        ,,^^  .  ^^^  ^  ^^  .  ^^  ^^ 

Now,  by  hypothesis  and  rule  of  equi-multiples, 

mA  :  mB  =  mX  :  mYj 
and  nC :  mB  =  nZ  :  mY.  [45 

Hence  mX:mY>  nZ'.mY\ 

therefore,  the  consequents  being  identical, 

mX  >  nZ.  [29 

Thus  the  second  pair  of  the  above  multiples  are  also  in 
descending  order  of  size. 


PROPORTION  269 

Next,  suppose  the  first  pair  to  be  in  ascending  order. 

The  student  may  treat  this  case  in  a  similar  way  ;  and  also  the 
remaining  case,  in  which  the  order  is  indifferent. 

It  follows  that  the  pairs  of  multiples 

mA,  nC 
and  mX,  nZ 

are  in  the  same  order  of  size,  whatever  m  and  n  are. 

Therefore  A  :  C  =  X :  Z. 

2.    Let  there  be  four  magnitudes  in  each  set ;   namely, 

A,  B,    C,  D 
and  X,  r,  z,   w. 

Let  the  successive  ratios  A  :  B,  B  :  C,  C  :  I)  he  respectively 
equal  to  the  successive  ratios  x  :  Y,  Y:  z,  Z  :  W. 
To  prove  that  the  ratios  of  the  extremes  are  equal, 

i.e.  A  :  D  =  X  :  W. 

The  student  may  prove  by  using  Case  1  twice  in  succession ;  and 
may  then  generalize. 

Applications  of  rule  of  succession. 

Equivalent  consequents  in  two  proportions. 

56.  Cor.  I.  If  there  are  two  proportions,  and  if  the  two 
consequents  in  one  proportion  are  equivalent  respectively  to  the 
two  consequents  in  the  other,  then  their  aiUecedents  form  a 
proportion. 

Given  A  :  B  =  X  :  Y, 

and  ^  ^' :  5  =  X' :  F; 

to  prove  A:  A'  =  X:  X\ 

Outline.    Compare  the  successive  ratios  in  the  two  sets 

A,  B,   A' 
and  X,  Y,  x'. 

Note.  Observe  that  the  rule  of  alternation  (47)  cannot  be  used  in 
proving  (56),  for  A  and  X  may  not  be  magnitudes  of  the  same  kind. 


270  PLANE  GEOMETRY  —  BOOK  IV 

Equivalent  antecedents  in  two  proportions. 

57.  Cor.  2.  If  there  are  two  proportions^  and  if  the  two 
antecedents  in  one  proportion  are  equivalent  respectively  to  the 
two  antecedents  in  the  other,  then  their  consequents  form  a 
proportion. 

Three  terms  equivalent  in  two  proportions. 

58.  Cor.  3.  If  two  proportions  have  any  three  terms  of  one 
equivalent  respectively  to  the  three  corresponding  terms  of  the 
other,  then  tJie  remaining  terms  are  equivalent. 

Rule  of  addition, 

59.  Theorem  17.  If  there  are  two  proportions,  and  if 
the  two  consequents  in  one  are  equivalent  respectively 
to  the  two  consequents  in  the  other,  then  the  sums 
of  corresponding  antecedents  form  a  proportion  with 
the  same  consequents. 

Given  A  :  B  =  x :  Y, 

and  A':B  =  X''.T', 

to  prove  A  -\-  A' :  B  =  X  -^  X' :  T. 

Since  the  two  given  proportions  have  their  consequents 
respectively  equivalent,  hence  their  antecedents  form  a  pro- 
portion ;  that  is,  a  :  A' =  X :  x',  [56 
therefore,  by  the  rule  of  composition, 

A-}-A':A'  =  X  +  X':X'',  [48 

now,  by  hypothesis, 

A' :  B  =  X' :  Y; 

therefore,  by  the  rule  of  succession,  • 

A-{-A':B=X-\-X':Y,  [55 

Rule  of  subtraction. 

60.  Cor.    In  the  same  case 

A  —  A':B  =  X—X':  Y. 
The  student  may  give  the  complete  statement  and  proof. 


BOOK  v.  — RATIOS   OF  LINES,   POLYGONS,  ETC. 

1.  The  general  principles  established  in  Book  IV  will 
now  be  used  in  comparing  particular  magnitudes  of  the  same 
kind,  —  chiefly  segments  of  lines  and  surfaces  of  polygons. 

SIMILARLY  DIVIDED  LINES 
Parallel  transversals, 

2.  Theorem  1.  If  two  lines  are  cut  hy  three  par- 
allels, any  two  of  the  intercepts  on  one  line  form 
a  proportion  with  the  corresponding  intercepts  on  the 
other  line. 

Let  the  lines  OL,  o'l'  be  cut  by  the  parallels  00',  A^A'^^ 
B^b\  making  the  three  pairs  of  corresponding  intercepts  OA^ 
and  oU'i,  OB^  and  0'b\,  A^B^^  and  J'iB\. 


1 

,     A2 

/gi^-- 

!     i  1 

1 
j 

f1 

1 

1 

/        / 

»                1 

I                1 

1 

I    i ! 

'    1 1 
i ! 

1    ! 

!       i 
1      i 
i       i 
i 

i 

A,  B,  A,        A^     B,A, 


First  to  prove    OA^ :  OBj^  =  0'A\:  0'b\. 

On  OL  lay  off  consecutive  segments  equal  to  OA^-,  and 
mark  their  extremities  with  the  symbols  A^,  A^,  •••.  Through 
these  points  draw  parallels  to  00'  meeting  o'l'  in  ^'2,  ^'g,---. 


271 


272  PLANE  GEOMETRY — BOOK  V 

Again,  on  OL  lay  off  consecutive  segments  equal  to  O^^; 
and  mark  their  extremities  B2,  B3,  •••.  Through  these 
points  draw  parallels  to  OO'  meeting  o'l'  in  B'2,  B'^,  •••. 

Any  segment  0A„  is  equal  to  the  mth  multiple  of  O^i; 
and  OBn  is  equal  to  the  ?ith  multiple  of  Ofij.  Hence  the 
scale  of  relation  of  OA^  and  OB^  is  as  shown  on  the  line 
OL.     (IV.  11.) 

The  parallels  through  A^,  A.2,  A^,  •••  make  equal  intercepts 
on  the  line  O'l'  (1. 167)  ;  thus  0'a'„  equals  the  mth  multiple 
of  0'A\.  Similarly,  O'b'^  equals  the  ;<th  multiple  of  0'b\.' 
Hence  the  scale  of  relation  of  0'A\  and  o'b\  is  as  shown 
on  the  line  O'l'. 

Since  like  multiples  evidently  occur  in  the  same  (frder  in 
the  two  scales,  hence  these  scales  of  relation  are  everywhere 
similar ;  and  it  follows,  by  definition  of  equal  ratios,  that 

OAi :  Oi^i  =  O'A'i :  O'b'^.  [IV.  15 

Next  to  prove  that 

OAi.AiBi=  O'A'i'.  A'^B'^. 
Since  OA^ :  OB^  =  O'a'^  :  o'b'^ 

therefore,  by  the  rule  of  separation, 

OAi  :0Bi—  OAi=  O'a'^  :  O'B'^  —  O'A'^,       [IV.  49 
i.e,        '  '  OA^:  A^Bi  =  O'A'i  :  A'^B'^. 

3.  Cor.  I.  If  two  lines  ore  cut  by  any  number  of  parallels 
the  segments  of  one  line  taken  in  order  as  antecedents  form  a 
series  of  equal  ratios  with  the  segments  of  the  other  line  taken 
in  order  as  consequents. 

For,  by  the  rule  of  alternation, 

0^1 :  O'A'^  =  Jjfii :  ^'iB'i=  •••.  [IV.  47 

Special  case. 

4.  Cor.  2.  If  two  sides  of  a  triangle  are  cut  by  parallels  to 
the  third  side,  the  segments  of  the  first  side  taken  in  order  as 
antecedents  form  a  seines  of  equal  ratios  ivith  the  segments  of 
the  second  side  taken  in  order  as  consequents. 


SIMILARLY  DIVIDED  LINES  273 

Definition.  Two  finite  lines  are  said  to  be  similarly 
divided  if  the  segments  of  the  first  taken  in  order  as  ante- 
cedents form  a  series  of  equal  ratios  with  the  segments  of 
the  second  taken  in  order  as  consequents. 

The  antecedent  and  consequent  of  any  one  of  the  equal 
ratios  are  called  corresponding  segments  of  the  two  lines. 

Two  points  of  division  are  said  to  correspond  if  the  two 
segments  adjacent  to  the  first  point  correspond  respectively 
to  the  two  segments  adjacent  to  the  second  point.  Two  end- 
points  are  said  to  correspond  if  the  segment  adjacent  to  one 
corresponds  to  the  segment  adjacent  to  the  other. 

Similar  Division 

5.  Problem  1.  To  divide  a  given  line  similarly  to 
a  given  divided  line,  tJiose  end-points  which  are  to 
correspond  being  started. 

Let  ^'^'  be  the  given  line  divided  at  the  points  P'  and  Q'; 
let  AB  be  the  other  given  line  which  it  is  required  to  divide 
similarly ;  and  let  A 
be  that  end-point 
which  is  to  corre- 
spond to  A\ 

Transfer    ^'J?'    so 
that  A^  may  fall  on 
the     corresponding 
point  A,  and  so  that  the  two  lines  may  form  a  convenient 
angle.     Join  B^B,  and  draw  P^P  and  Q'Q  parallel  to  B^B. 

Then,  from  the  principle  of  parallel  transversals, 

AP:  A'P^  =  PQ:P'Q^=QB:  Q'b' ;  [3,4 

therefore  the  lines  AB  and  A'b'  are  similarly  divided. 

6.  Cor.  I.  WJien  two  lines  are  similarly  divided,  the  ratio 
of  tivo  corresponding  segments  equals  the  ratio  of  the  whole  lines. 

[In  a  set  of  equal  ratios  any  antecedent  is  to  its  consequent  as  the 
sum  of  the  antecedents  is  to  the  sum  of  the  consequents  (IV.  53).] 


274  PLANE  GEOMETRY  —  BOOK   V 

7.  Cor.  2.  When  two  lines  are  similarly  divided,  the  ratio 
of  two  corresponding  segments  is  the  same  whatever  he  the 
mode  of  division. 

8.  Discussion  of  Problem  1.  It  follows  from  7  that  when 
any  segment  PQ  of  the  first  line  is  given,  then  the  corre- 

•    sponding   segment  P'Q^  of  the   second   line  is   'uniquely 
determined.'     Therefore  there  is  only  one  solution. 

Ex.  Divide  a  given  line  into  three  parts  so  that  the  ratios  of  the 
parts  may  equal  the  ratios  of  three  given  lines,  or  of  given  numbers. 

Converse  of  2, 

9.  Theorem  2.  If  two  similarly  divided  lines  are 
plaxied  so  that  tJie  li?ie  joining  one  pair  of  corre- 
sponding points  is  parallel  to  the  line  joining  an- 
other pair  of  corresponding  points,  then  the  lines 
joining  all  pairs  of  corresponding  points  are  parallel. 

Let  the  line  OP  be  divided  at  the  points  A  and  B  ;  and  let 
O^P^  be  divided  similarly  at  A^  and  B\  in  such  a  way  that  0 
corresponds  to  0\  A  \,o  A\B  to 
b\  p  to  P\    Let  00^  be  par-  /^^^^  ^/ 

allel  to  PP\ 

To  prove  that  OQ',  AA\  and 
55'  are  parallel  to  each  other. 

Suppose,  if  possible,  that 
they  are  not  all  parallel ;  and 
let  lines  be  drawn  through  A  and  B  parallel  to  00^  and  PP\ 

These  parallels  will  divide  O^P^  similarly  to  OP ;  and  will 
therefore  pass  through  the  points  A '  and  B' ;  because  there  is 
only  one  way  of  dividing  the  line  O'P'  similarly  to  OP  (8). 

Hence  A  A'  and  BB'  are  parallel  to  00'  and  PP'. 

Converse  of  4, 

10.  Cor.  If  t2co  similarly  divided  lines  are  placed  so  as  to 
have  two  corresponding  points  in  coincidence,  then  the  lines 
joining  the  other  corresponding  points  are  parallel. 


SIMILARLY  DIVIDED  LINES  275 


DIVISION    IN   A    GIVEN    RATIO 

11.  Definition.  A  ratio  is  said  to  be  given  when  its  ante- 
cedent and  consequent  are  given  magnitudes.  Usually  the 
most  convenient  magnitudes  by  which  a  ratio  can  be  as- 
signed are  either  line-segments  or  whole  numbers.  The 
latter  can,  however,  express  only  the  ratio  of  commensur- 
able magnitudes. 

A  line  is  said  to  be  divided  in  a  given  ratio  at  a  point  when 
the  ratio  of  the  two  segments  is  equal  to  the  given  ratio. 

Internal  division, 

12.  Problem  2.  To  divide  a  given  line  internally 
into  two  segments  whose  ratio  shall  Ibe  equal  to  a 
given  ratio ;  that  end  of  the  line  to  which  the  antece- 
dent is  to  he  adjacent  being  stated. 

Let  AB  be  the  given  line ; 
and  let  L  and  i»f  be  the  ante- 
cedent and  consequent  of  the 
given  ratio. 

To  find  a  point  P  in  the 
line  AB  such  that 

AP  •.PB=L\M. 

Draw  AD,  making  a  convenient  angle  with  AB.  Lay  off 
AC  equal  to  L,  and  CD  equal  to  M.  Join  z>5;  and  draw  CP 
parallel  to  DB. 

Then,  from  the  principle  of  parallel  transversals, 

AP  :PB  —  L:  M.  [4 

Show  that  there  is  only  one  sokition.     (8.) 
Ex.  1.    Divide  a  given  line  internally  in  the  ratio  2  :  3. 
Ex.  2.    Find  two  lines  whose  ratio  shall  be  equal  to  a  given  ratio 
and  whose  sum  shall  be  equal  to  a  given  line. 

Ex.  3.    Divide  a  line  into  three  parts  in  the  ratios  2:3:4. 


L 

D 

M 

^^ 

c.--""\ 

y 

"^\  \ 

276  PLANE  GEOMETRY  —  BOOK  V 

External  division, 

13.  Cor.  To  divide  a  given  line  externally  so  that  the  two 
segments  may  have  a  given  ratio ;  that  end  of  the  line  to  which 
the  antecedent  is  to  be  adjaceiit  being  stated. 

The  construction  is  similar 

to  12.    Show  that  there  is  only      — 

one  solution  ;  and  that  there  is  C 

no  solution  when  L  and  M  are      M. —  ,,'''  \^ 

equal.  .<'' 

Ex.  1.  Find  two  lines  whose  ,,''"'       \  \ 

ratio  shall  be  equal  to  a  given  ,.-'' 

ratio  and  whose  difference  shall     j^ ^ ^ 

be  equal  to  a  given  line. 

Ex.  2.  Divide  a  given  line  externally  in  the  ratio  3  : 2. 

CONVERSION    OF    LINE-RATIOS 

Arts.  14-21  will  treat  of  several  problems  in  the  construc- 
tion of  ratios  subject  to  assigned  conditions.  These  problems 
are  classified  under  the  heading  "  Conversion  of  line-ratios," 
which  will  be  explained  in  Art.  15.  They  are  solved  by 
the  principle  of  "  similar  division,"  which  has  been  exempli- 
fied in  the  preceding  articles. 

Fourth  proportional, 

14.  Problem  3.  To  find  a  fourth  proportional  to 
three  given  lines. 

Let  L,  Mj  and  N  be  the  three  given  lines. 
To  find  a  fourth   line,  P, 
such  that 


^  «.- 


L  :  M  =  N  :  P.  jy 

G'^'^'P     \ 
Place  two  indefinite   lines  ,--  \ 

OB  and  OD  at  any  convenient  ,,--'' 

angle.     Lay  off  OA  equal  to     q'--- jj-.4f_...>^. 

L,  AB  equal  to  M,  OC  equal 

to  N.     Join  AC ;  and  draw  BD  parallel  to  AC. 


SIMILARLY  DIVIDED  LINES  277 

The  line  CD  is  the  required  fourth  proportional. 

Since  OA:  AB  =  OC  :  CD,  [2 

therefore,  CD  is  a   fourth  proportional  to  the  three  lines 
OA,  AB,  and  OC ;  that  is,  to  the  three  given  lines  L,  M,  and  N. 

Discussion.     All  the  fourth  proportionals  to  three  given 
magnitudes  are  equivalent  (IV.  58). 

Therefore  there  is  only  one  solution  to  this  problem. 

Note.     Problem  3  may  also  be  stated  thus : 

Given  a  line  N,  to  find  another  line  P  such  that  the  ratio  of 
NtoP  may  he  equal  to  a  given  ratio; 

or  thus : 

Given  the  ratio  of  two  lines,  and  given  the  antecedent,  find 
the  consequent. 

Ex.  1.    Given  a  line  N,  con-  ,'-^, 

struct  another  line  P  such  that 


iV^:P=3:4. 

Ex.  2.     Given  a  line  P,  con- 
struct another  line  N  such  that 

iV^:P=3:4. 


N 


15.  Definition.     To  convert  a  given  ratio  is  to  find  an 
equal  ratio  so  as  to  satisfy  stated  conditions. 

E.g.,  problem  3  may  be  enunciated  thus : 

To  convert  a  given  ratio  {L  :  M)  so  that  the  new  antecedent 
may  equal  a  given  line  (iV). 

16.  Cor.  1.     Convert  the  line-ratio  L:  M  so  that  the  new 
consequent  may  equal  P. 

Lay  off  M,  L,  P  as  in  figure,  ^,,''''    \^ 

and  construct  as  before.     Then        ,.--''''    '"-. 


N  is  the  required  antecedent.  M  L 

Ex.    Given  a  numerical  ratio  m:n,  convert  it  into  a  line-ratio 
whose  consequent  shall  be  a  given  line  (see  14,  ex.  2). 


278  PLANE  GEOMETRY — BOOK  V 

17.  Cor.  2.  Given  any  two  line-ratioSf  convert  them  so  as 
to  have  a  common  consequent. 

Note.  By  this  method  two  ratios  can  be  compared  with  each  other, 
so  as  to  determine  whether  the  first  is  greater  than,  equal  to,  or  less 
than  the  second. 

Ex.  Convert  the  ratios  3  : 4  and  4  :  5  so  as  to  have  a  common  con- 
sequent (IV.  36),  and  then  show  which  ratio  is  the  greater. 

18.  Definition.  To  enlarge  or  reduce  a  given  line  in  a 
given  ratio  is  to  find  another  line  such  that  the  given  line  is 
to  the  new  line  in  the  given  ratio. 

E.g.^  the  line  P  in  14,  ex.  1,  is  an  enlargement  of  N  in  the  ratio 
3:4. 

An  enlargement  or  reduction  is  called  an  alteration. 
E.g..,  the  line  P  in  14  is  an  alteration  of  N  in  the  ratio  L  :  M. 

Ex.  ^To  reduce  a  given  line  OA  in 
the  ratio  6  :  2. 

Here  OA  :  OA'  =  6:2.  Show  also 
that  A' A  is  an  enlargement  of  OA'  in 
the  ratio  of  2  :  3. 

19.  Definition.  Any  set  of  magnitudes  of  the  same  kind 
are  said  to  be  in  continued  proportion  when  the  successive 
ratios  of  the  set  are  all  equal  (IV.  54).  When  three  mag- 
nitudes are  in  continued  proportion,  the  third  magnitude  is 
said  to  be  a  third  proportional  to  the  first  and  second; 
and  the  second  is  said  to  be  a  mean  proportional  between 
the  first  and  third. 

Third  proportionaL 

20.  Problem  4.  To  find  a  third  proportional  to  two 
given  lines. 

Let  L  and  M  be  the  two  given  lines. 
To  find  a  third  line  N  such  that 

L'.M  =  M'.N. 

[Find  a  fourth  proportional  to  i,  M^  and  N,  by  means  of  14.] 


COMPOUNDING   OF  RATIOS  ■  279 

21.  Cor.  Given  two  lines  L  and  M ;  find  N  and  P,  such  that 
L,  M,  N,  and  P  may  form  a  continued  proportion. 

Ex.  Find  a  third  proportional  to  the  numbers  2  and  6.  Continue 
this  proportion  for  two  terms  more. 

COMPOUNDING  OF  RATIOS 

22.  Definition.  If  there  is  any  set  of  like  magnitudes, 
the  first  is  said  to  have  to  the  last  the  ratio  compounded 
of  the  successive  ratios  of  the  set  (see  IV.  54). 

E.g.,  if  there  are  four  like  magnitudes  A,  B,  c,  D,  then  the 
ratio  of  the  extremes  A:  D  i^  compounded  of  the  successive 
ratios  a:B,  B  :  C,  C  :  D. 

If  there  is  any  number  of  given  ratios,  whether  successive 
or  not,  and  if  there  is  found  a  set  of  magnitudes  whose 
successive  ratios  are  respectively  equal  to  the  given  ratios, 
then  the  extremes  of  this  set  are  said  to  have  a  ratio  com- 
pounded of  the  given  ratios. 

E.g.,  if  there  are  any  three  ratios 

A:B,  M:  N,  X:  Y] 
and  if  there  are  found  any  four  like  magnitudes 

P,  Q,  R,  S, 
such  that  their  three  successive  ratios  are  respectively  equal 
to  the  given  ratios,  then  the  ratio  of  their  extremes,  P :  S 
(which  by  the  preceding  definition  is  compounded  of   the 
three  successive  ratios 

P.Q,    Q:R,  R:S), 
is  by  the  present  definition  also  said  to  be  compounded  of 
the  respectively  equal  ratios 

A:  B,  M:  N,  X  :  Y. 

The  set  of  magnitudes  just  mentioned  whose  successive 
ratios  are  respectively  equal  to  the  given  ratios  are  called 
aujciliary  jnagnitudes  to  the  given  ratios. 

The  use  of  auxiliary  magnitudes  is  illustrated  in  the  next 
theorem. 

MOM.    ELEM.    GEOM.  —  19 


280  PLANE  GEOMETRY  —  BOOK  V 

Fundamental  principle  in  cotnpounding  ratios, 

23.  Theorem  3.  If  there  is  any  set  of  ratios  and 
another  set  severally  equal  to  them,  then  the  ratio 
compounded  of  the  first  set  is  equal  to  the  ratio  com^- 
pounded  of  the  second  set. 

Let  the  ratios  of  the  first  set  be 

A  :  Bj  P  :  Qj  X  :  T, 

and  those  of  the  second  set 

A'  :  B',  P' :  Q',  X' :  7', 

the  former  ratios  being  respectively  equal  to  the  latter. 

To  prove  that  the  ratio  compounded  of  the  first  set  is 
equal  to  that  compounded  of  the  second  set. 

Take  an  auxiliary  set  of  lines 

F,  (?,  H,  K, 

such  that  their  successive  ratios  are  respectively  equal  to 
the  ratios  of  the  first  set. 

[This  may  be  done  by  taking  an  arbitrary  line  -F;  then 
finding  0  so  that  A  :  B  =  F  :  0  ;  next  finding  H  so  that  P  :  Q 
=  G  :  Hy  and  so  on  (14).] 

Similarly  take  a  set  of  lines 

F'y   (?',  H\   K\ 

such  that  their  successive  ratios  are  respectively  equal  to 
the  ratios  of  the  second  set. 

Then,  by  hypothesis  and  IV.  19,  the  successive  ratios  of  the 
set  Fj  G,  H,  K  are  respectively  equal  to  the  successive  ratios 
of  the  set  F',  G',  H\  K'. 

Hence  the  ratios  of  their  extremes  are  equal  (IV.  55); 
that  is,  F  :  K  =  F'  :  K'. 

But  F  :  K  is  by  definition  the  ratio  compounded  of  the 
ratios  of  the  first  set ;  and  F' :  K'  is  the  ratio  compounded 
of  the  ratios  of  the  second  set. 

Hence  the  theorem  is  established. 


COMPOUNDING   OF  RATIOS  281 

24.  Note.  It  follows  from  this  fundamental  theorem 
that  in  compounding  any  given  ratios  it  makes  no  difference 
what  auxiliary  magnitudes  are  chosen  provided  their  suc- 
cessive ratios  are  respectively  equal  to  the  given  ratios. 

The  operation  may  be  conveniently  regarded  as  the  per- 
formance of  several  successive  'alterations'  (18).  If  there 
are  any  magnitudes  of  the  same  kind  L,  M,  N,  P,  the  successive 
operations  of  altering  L  to  M,  M  to  N,  N  to  P,  give  the  same 
result  as  altering  L  directly  to  P.  Thus  the  ratio  L  :  P  is 
appropriately  said  to  be  compounded  of  the  successive  ratios 
L  :  M,  M :  N,  N  :  P,  or  oi  any  three  ratios  equal  to  these. 

Hence  any  given  ratios  a:b,  c:d,  e-.f  may  be  compounded 
as  follows : 

Assume  any  line  L^  alter  it  to  M  in  the  ratio  a :  b,  alter  M 
to  N  in  the  ratio  c :  d,  alter  iv  to  P  in  the  ratio  e :  /,  then 
L  :  P  is  equal  to  the  ratio  compounded  of  the  three  ratios 
a:b,  c:  d,  e:f. 

Compounding  line-ratios. 

25.  Cor.  I.     To  give  a 

simple    construction    for  ^^^'^^^'^ 

compounding     the     line-  a^^^\  I 

ratios  a:b,  c:  d  into  one    ^^^-^^      .'      m      I  N 

ratio  L  :  N  whose  effect 
shall  be  the  same  as  their 
joint  effects. 

Compounding  numerical  ratios, 

26.  Cor.  2.  If  the  terms  of  two  ratios  are  numbers,  the  ratio 
compounded  of  them  is  equal  to  the  ratio  whose  antecedent  is 
the  product  of  their  antecedents  and  whose  consequent  is  the 
product  of  their  consequents. 

[Show  that  the  ratios  m  :  n  and  p  :  q  have  the  auxiliary  magnitudes 

mp,  np,  ng.] 

Ex.    Show  that  the  ratio  compounded  of  ^  :  J5  and  m  :  w  is  mA  :  nB. 
[The  auxiliary  magnitudes  are  mA,  mB,  nB  (IV.  35,  38).] 


282  PLANE  GEOMETRY— BOOK   V 

Order  of  compounding, 

27.  Theorem  4.  The  order  in  which  two  given 
ratios  are  compounded  is  indifferent. 

Let  there  be  two  ratios  A\  B^  X  :  Y. 

To  prove  that  the  ratio  obtained  by  compounding  them  is 
the  same  in  whichever  order  they  be  taken. 

Take  any  line  L.  Alter  L  to  3/  in  the  ratio  A:  B.  Alter 
M  to  N  in  the  ratio  X :  Y.  Then  the  two  ratios  (taken  in 
the  order  named)  have  the  auxiliary  magnitudes 

L,        3f,        Ny 

and  hence  compound  into  the  ratio  L  :  N. 

Next  take  the  two  ratios  in  the  order  X  :  r,  A:  B. 

Take  any  line  L\  Alter  L'  to  M^  in  the  ratio  X  :  Y.  Alter 
3f'  to  JV'  in  the  ratio  A:  B.  Then  the  two  ratios  (taken  in 
this  order)  have  the  auxiliary  magnitudes 

L\  M',  N', 
and  hence  compound  into  the  ratio  L' :  N'. 

Now  it  is  to  be  proved  that 

L:N  =  L':N'. 

Find  P  a  fourth  proportional  such  that  L  :  M  =  N :  P, 

Then  the  two  sets,       m   N    P 
and  L\  M\  N\ 

have  their  successive  ratios  respectively  equal, 

for  L^'.M^  =  X'.Y=  M'.Nj 

and  M* :  N'  =  A  :  B  =  L  :  M  =  N :  P. 

Hence  the  extremes  of  the  two  sets  are  proportional, 
i.e.  M:P=L':N';  [IV.  55 

now  M :  P  =  L :  N,^  [alternation 

therefore  L:N  =  L':N',  [IV.  19 

Hence  the  theorem  is  proved. 

Ex.    The  order  of  compounding  three  ratios  is  indifferent. 


COMPOUNDING   OF  RATIOS  283 

Equal  ratios  compounded  tvith  unequal  ratios. 

28.  Theorem  5.  If  one  ratio  is  greater  than  another, 
then  the  ratio  compounded  of  the  greater  and  any 
third  ratio  is  greater  than  that  compounded  of  the. 
less  and  the  same  third  ratio. 

Let  A\B  >  P:Q, 

and  let  X :  r  be  any  other  ratio. 

To  prove  that  the  ratio  compounded  of 
A  :  B  and  X  :  Y 
is  greater  than  the  ratio  compounded  of 
P  :  Q  and  X :  Y. 

Take  any  line  L.  Alter  it  to  M  in  the  ratio  X:  Y.  Alter 
M  to  N  in  the  ratio  A:  B. 

Then  the  ratios  X :  Y  and  A :  B  have  the  auxiliary  magni- 
t"<ies  jr^  M^  ^,  |-22,  def. 

Again  alter  3f  to  iV^'  in  the  ratio  P  :  Q. 
Then  the  ratios  X :  Y  and  P  :  Q  have  the  auxiliary  niagni- 
t^^des  L,M,N'. 

It  is  now  to  be  proved  that 

L:N  >  L:N'. 

Since  A:B  >  P  :  Q^  V^JV- 

therefore  M:N  >  M:N\ 

hence  .v'  >  N,  [IV.  32 

therefore  L:N>  L:N'.  [IV.  30 

Hence  the  required  result  is  proved. 

29.  Cor.     If  A.B  >P:Q 
and                                     L:M>  X:Y, 

then  the  ratio  compounded  of  A:  B  and  L:  M  is  greater  than 
that  compounded  of  P  :  Q  and  X:Y. 
[Apply  the  theorem  twice.] 


284  PLANE  GEOMETRY — BOOK  V 

Duplication  of  a  ratio, 

30.  Definition.  When  two  ratios  are  equal,  the  ratio  com- 
pounded of  them  is  called  the  duplicate  of  either  of  them. 

When  three  ratios  are  equal,  the  ratio  compounded  of  them 
is  called  the  triplicate  of  any  one  of  the  original  ratios. 

31.  Theorem  6.  //  three  magnitudes  are  propor- 
tional, then  the  ratio  of  the  first  to  the  third  is  equal 
to  the  duplicate  of  the  ratio  of  the  first  to  the  second. 

Given  A.  B  =  5  :  C ; 

to  prove  A:C  —  duplicate  oi  A:B. 

The  ratio  of  ^ :  C  is  compounded  of  the  successive  ratios 
A  :  B  and  B  :  C  (22,  def.).     But  these  two  ratios  are  equal. 

Therefore  the  ratio  compounded  of  them  is  the  duplicate  of 
either  (30,  def.).    Hence  A  :  C  equals  the  duplicate  of  A:  B. 

32.  Cor.  I.  If  four  magnitudes  are  in  continued  proportion  J 
the  ratio  of  the  first  to  the  fotirth  is  equal  to  the  triplicate  of  the 
ratio  of  the  first  to  the  second. 

33.  Cor.  2.  To  find  a  ratio  equal  to  the  duplicate  of  a  given 
line-ratio  ;  also  of  a  given  numerical  ratio. 

34.  Cor.  3.     To  find  the  triplicate  of  a  given  line-ratio. 

Comparison  of  duplicate  ratios, 

35.  Theorem  7.  According  as  one  ratio  is  greater 
than,  equal  to,  or  less  than  another,  so  is  the  duplicate 
of  the  former  greater  than,  equal  to,  or  less  than  the 
duplicate  of  the  latter. 

If  A-.B^P'.Qy 

the  ratio  compounded  oi  A:  B  and  A:B  is  equal  to  the  ratio 
compounded  of  P  :  Q  and  P :  Q  (23). 

If  A:B>P:Q, 

the  ratio  compounded  oi  A:B  and  A:B  i?,  greater  than  that 
compounded  of  P  :  Q  and  P  :  Q  (29). 

Hence  the  theorem  is  established. 


SIMILAR   TRIANGLES  285 

36.  Cor.  One  ratio  is  greater  than,  equal  to,  or  less  than 
another  according  as  the  duplicate  of  the  first  ratio  is  greater 
than,  equal  to,  or  less  than  the  duplicate  of  the  second. 

SIMILAR  TRIANGLES 

This  section  and  the  next  will  treat  of  similar  triangles 
and  similar  polygons,  respectively.  In  the  following  general 
definitions  the  word  "  polygon  "  will  be  understood  to  include 
"triangle."  A  former  definition  is  here  repeated  for  con- 
venience. 

37.  Definitions.  Two  polygons  are  said  to  be  mutually 
equiangular  if  the  angles  of  one,  taken  in  order,  are  equal 
respectively  to  those  of  the  other  taken  in  order.  The  equal 
angles  are  said  to  correspond;  and  the  sides  joining  the 
vertices  of  corresponding  angles  are  called  corresponding 
sides. 

Two  polygons  are  said  to  have  their  sides  proportional 
if  the  sides  of  one,  taken  in  order  as  antecedents,  form  a 
series  of  equal  ratios  with  the  sides  of  the  other  taken  in 
order  as  consequents. 

Two  polygons  are  said  to  be  similar  if  they  are  mutually 
equiangular,  and  if  the  corresponding  sides  are  proportional. 

The  ratio  of  any  two  corresponding  sides  is  called  the 
ratio  of  similitude  of  the  similar  polygons. 

E.g.,  the  quadrangles  ABCD  and  a'b'c'd'  are  similar  if 
the  angles  A,  B,  C,  D  are  equal  respectively  to  A\  B\  C',  D', 
and  if       AB  :  A'b'  =  BC  :  B'c'  =  CD  :  C'd'  =  DA  :  D'A'. 

Each  of  these  ratios  is  equal  to  the  ratio  of  similitude  of 
the  similar  quadrangles. 

Two  similar  polygons  are  said  to  be  directly  or  ohversely 
similar  according  as  they  are  directly  or  obversely  equi- 
angular (I.  187). 

Ex.     Two  regular  polygons  of  the  same  number  of  sides  are  similar. 


286  PLANE  GEOMETRY — BOOK  V 

Conditions  op  Similarity 

The  next  four  theorems  relate  to  the  conditions  of  simi- 
larity of  two  triangles. 

Angles  equal, 

38.  Theorem  8.  If  two  triangles  are  mutually 
equiangular,  then  their  sides  are  proportional;  and 
the  triangles  are  similar. 

Let  the  triangles  ABC  and  A'b'c'  be  equiangular. 

C 


A' 
To  prove  that 

AB  :  A'B^  =  BC  :  5'c'  =  CA  :  &A\ 
Apply  the  triangle  A' B^c' to  ABC  so  that  A^  coincides 
with  A,  and  ^'5'  falls  on  ^s  ;  then  A^c'  falls  on  AC^  because 
the  angles  A  and  A^  are  equal.  Let  B'  and  cf  take  the 
respective  positions  iJ"  and  c"  on  the  sides  AB  and  ^c  or 
else  on  their  prolongations. 

Since  the  angles  B  and  B"  are  equal,  the  lines  BC  and 
5"C"  are  parallel;   therefore,  by  theorem  1, 

AB  :  AB"  =  AC :  AC"y 
i.e.  AB  :  A'B'  =  AC  :  A'c'. 

Similarly  by  applying  the  angle  B'  to  the  angle  B  it  may 
be  shown  that  aB:A'b'  =  BC:  B'c'. 

39.  Cor.  I.     A  parallel  to  one  side  of  a  triangle  foims  with 
the  other  two  sides  a  similar  triangle. 

40.  Cor.  2.     Triangles  whose  sides  are  parallel,  respectively  J 
are  similar. 


SIMILAR    TRIANGLES  287 

Construction  of  similar  triangles, 

41.  Problem  5.  To  construct  a  triangle  similar  to 
a  given  one,  arid  such  that  the  ratio  of  similitude  is 
equal  to  a  given  ratio. 

Let  ABC  be  the  given  triangle,  and  L  :  M  the  given  ratio. 


/ 

/' 
/' 

TA"' 
\ 

\    . 

-::^^'' 

,^:'- 

A' 

~'yB' 

M 


To  construct  a  triangle  A^B^&  similar  to  ABC,  and  such 
*^^*  AB  :  A'B^  =  L:  M. 

Take  any  point  0  ;  and  draw  OA,  OB,  OC.     Find  the  fourth 
proportional  to  L,  M,  and  OA  (14).     Lay  off  OA'  equal  to 
this  fourth  proportional.     Draw  A'b'  parallel  to  AB,  and 
A'c'  parallel  to  AC;  and  join  B'c'. 
Then  A'b'c'  is  the  required  triangle. 
Since  AB  is  parallel  to  A'b',  hence 

OA  :  OA' =  OB  :  OB' ',  [2 

and  since  ^C  is  parallel  to  B'c',  then 

OA:  OA'  =  OC:  OC'. 
Therefore,  by  equality  of  ratios, 

OB  :  OB'  =  OC:  OC'.  [IV.  19 

Hence  BC  is  parallel  to  B'c'.  [10 

Therefore  the  triangles  ABC  and  A'b'c'  are  similar. 
Moreover,  their  ratio  of  similitude,  AB  :  A' B',  is  equal  to 

OA  :  OA',  and  is  therefore  equal,  by  construction,  to   the 

given  ratio  L  :  M. 


288  PLANE  GEOMETRY — BOOK  V 

42.  Definition.  The  new  triangle  is  called  a  reduction 
or  enlargement  of  the  given  one,  according  as  the  conse- 
quent of  the  given  ratio  is  less  or  greater  than  the  antecedent. 

Two  similar  triangles  (or  polygons)  are  said  to  have  the 
same  shape  or  pattern.  Thus,  in  a  reduction  or  enlarge- 
ment, the  size  is  altered,  but  the  shape  is  preserved. 

Sides  proportional, 

43.  Theorem  9.  If  two  triangles  have  their  sides 
proportional,  then  they  are  Tiiutually  equiangular, 
and  the  triangles  are  similar. 

Let  the  triangles  ABC  and  A'B^&  have  their  sides  such 

that  AB  :A'b' =  BCB'C'^  ^'  ■  ^'" 


To  prove  that  the  triangles  are  similar. 

Lay  off  AB"  equal  to  A'b';  and  draw  B"c"  parallel  to  BC. 

The  triangles  ABC  and  AB"c"  are  mutually  equiangular. 

Therefore,  AB  :  AB"  =  AC :  AC";  [38 

now  A'B:A'B'  =  AC:A'c'f  [hyp. 

and  AB"  =  a'b',  [constr. 

hence  the  two  proportions  have  three  corresponding  terms 
respectively  equal ; 

therefore,  AC"  =  A'c'.  [IV.  58 

Similarly,  it  may  be  proved  that 
B"c"  =  B'c'. 

Hence  the  triangle  a'b'c'  is  equal  to  AB"c"f  and  there- 
fore similar  to  ABC. 


SIMILAR   TRIANGLES  289 

Two  sides  and  included  angle, 

44.  Theorem  10.  If  two  triangles  have  an  angle 
of  one  equal  to  an  angle  of  the  other,  and  the  sides 
ahoiAjt  these  angles  proportional,  then  the  triangles 
are  similar. 

Let  the  triaDgles  ABC  and  A^B^&  have  the  angles  A  and 
A^  equal,  and  also  have 

AB  :A'B^  =  AC:  A'C\ 

G 


A'  B      A 

To  prove  the  triangles  similar. 

Lay  off  ^^"  equal  to  ^'j5' ;  and  draw  jB"c"  parallel  to  BC. 

The  triangles  ABC  and  AB"c"  are  similar  (38). 

Therefore,  AB  :  AB"  =  AC  :  Ac"  ; 

now  AB  :A'b'  =  AC  lA'C'y  [hyp. 

and  AB"  =  A'b',  [constr. 

therefore,  AC"  =  A'c'.  [IV.  58 

Hence  the  triangle  A'b'C'  equals  AB"c"j  and  is  therefore 
similar  to  ABC. 

Two  sides  and  a  non-included  angle, 

45.  Theorem  11.  If  two  triangles  have  an  angle  of 
one  equal  to  an  angle  of  the  other,  and  if  the  sides 
about  another  angle  in  each  are  proportional  {in 
such  a  way  that  the  sides  opposite  the  equal  angles 
correspond),  then  the  third  angles  are  either  squal  or 
supplemental. 

Let  the  triangles  ABC  and  a'b'c'  have  the  angles  B  and 
B'  equal,  and  have  the  sides  about  the  angles  A  and  A'  pro- 
portional such  that     AB  :  A'b'  =z  AC  :  A  'C'. 


290 


PLANE  GEOMETRY  —  BOOK  V 


To  prove  that  the  angles  C  and  c'  are  either  equal  or 
supplemental. 

The  angles  A  and  A' 
included  by  the  propor- 
tional sides  are  either 
equal  or  unequal. 

If  these  angles  are 
equal,  as  in  figure  1,  the 
third  angles  C  and  c'  are 
equal  (I.  130). 

If  the  angles  A  and  A^ 
are  unequal,  let  A  be  the 
greater,  as  in  figure  2. 

Draw  AC'^  cutting  off 
from  A  a  part  5^  C"  equal 
to^'. 

The  triangles  BAC"  and  b'A'c'  are  mutually  equiangular, 
since  they  have  two  angles  of  one  equal  to  two  angles  of  the 
other. 


Therefore 

AB:AC''  =  A'li' 

:A'C'', 

[38 

now 

AB'.AC    =A'b' 

:A'C'; 

[hyp. 

therefore 

AC"  =AC. 

[IV.  58 

Hence  the  angle  C  equals  AC"C,  which  equals  the  supple- 
ment of  AC"B.  Now  ^c"J5  equals  C';  therefore  the  angle 
C  equals  the  supplement  of  C'. 

Ex.  1.  Summarize  the  four  conditions  under  which  two  triangles 
are  similar.  Compare  them  with  the  five  conditions  under  which  two 
triangles  are  equal.  What  two  of  the  latter  correspond  to  the  first  of 
the  former  ? 

Ex.  2.  The  ratio  of  the  perimeters  of  two  similar  triangles  is  equal 
to  their  ratio  of  similitude.     (Use  IV.  53.) 

Ex.  3.  Prove  by  the  principle  of  similarity  that  the  line  joining  the 
mid-points  of  two  sides  of  a  triangle  is  parallel  to  the  third  side,  and 
equal  to  half  of  it. 


SIMILAR   TRIANGLES  291 

Applications  op  Similar  Triangles 

Might  triangle  divisible  into  similar  parts o 

46.  Theorem  12.  In  a  right  triangle,  the  perpen- 
dicular from  the  vertex  of  the  right  angle  to  the 
hypotenuse  divides  the  triangle  into  two  parts  sim- 
ilar to  the  whole  and  to  each  other. 

Let  ABC  he,  a  triangle,  right  angled  at  c,  and  let  GD  be 
perpendicular  to  AB.  ^ 

To  prove  that  the  triangles  ACD 
and  CBD  are  similar  to  ABC  and  to 
each  other. 

[Prove  the  three  triangles  equiangular  ;    ^  ^ 

the  corresponding  angles  being  those  marked  in  figure.] 

47.  Note.  The  corresponding  sides  are  opposite  equal 
angles.  In  the  triangles  ACD  and  CBD,  the  side  AD  of  the 
first  corresponds  to  CD  of  the  second,  and  the  side  CD  of  the 
first  corresponds  to  DB  of  the  second. 

The  ratio  of  similitude  of  these  two  triangles  equals  either 
of  the  ratios  AD  :  CD,  CD  :  DB,  AC:  CB. 

MEAN    proportional 

48.  Cor.  I.  In  a  right  triangle  the  perpendicular  on  the 
hypotenuse  is  a  mean  proportional  between  the  segments  of  the 
hypotenuse.     (19,  def.) 

49.  Cor.  2.  Conversely,  if  the  perpendicular  from  the  vertex 
to  the  base  is  a  mean  proportional  between  the  segments  of  the 
base,  then  the  vertical  ayigle  is  a  right  angle. 

50.  Cor.  3.  In  a  right  triangle  either  side  is  a  mean  pro- 
portional between  the  hypotenuse  and.  the  adjacent  segment  of 
the  hypotenuse  made  by  the  perpendicular. 

[Compare  corresponding  sides  of  the  triangles  ABC  and  ACD.'] 


292  PLANE  GEOMETRY — BOOK  V 

51.  Cor.  4.  The  segments  of  the  hypotenuse  are  in  the 
duplicate  ratio  of  the  two  sides. 

Outline.  The  ratio  AD  :  J)B  is  compounded  of  the  ratios 
AD  :  CD  and  CD  :  DB,  by  definition.  Now  each  of  these  equals 
AC :  CB ;  and  the  ratio  compounded  of  a  ratio  and  itself  is 
its  duplicate  ratio  (30).     Therefore,  etc. 

Ex,  1.  By  means  of  theorem  12  find  a  third  proportional  to  two 
given  lines. 

[Let  AD  and  DC  he  the  two  lines.] 

Ex.  2.  If  in  a  right  triangle  one  of  the  sides  is  double  the  other,  in 
what  ratio  does  the  perpendicular  divide  the  hypotenuse  ? 

Ex.  "3.  A  perpendicular  drawn  from  any  point  of  a  circle  to  a 
diameter  is  a  mean  proportional  between  the  segments  of  the  diameter. 

Ex.  4.  The  radius  of  a  circle  is  a  mean  proportional  between  the 
segments  of  a  tangent  between  the  point  of  contact  and  any  pair  of 
parallel  tangents. 

Construction  of  mean  proportioned, 

52.  Problem  6.  To  find  a  mean  proportional  he- 
tween  two  given  lines. 


Afj 


N 


N 


Use  theorem  12,  cor.  1,  and  ex.  3. 

Show  that  there  is  only  one  mean  proportional.     (Use  49.) 

Ex.  1.  Give  another  construction  by  taking  AB  equal  to  the  greater 
of  the  two  given  lines  and  AD  equal  to  the  less,  and  then  using 
theorem  12,  cor.  3. 

Ex.  2.  To  find  a  ratio  whose  duplicate  shall  be  equal  to  a  given 
ratio.     Show  that  there  is  only  one  solution. 

Ex.  3.  Show  that  the  mean  proportional  between  two  unequal 
lines  is  less  than  half  their  sum. 


SIMILAR   TRIANGLES 


293 


Division  of  base  by  angle-bisector, 

53.  Theorem  13.  If  the  interior  or  exterior  vertical 
angle  of  a  triangle  is  bisected  by  a  line  which  cuts 
the   base,    then   the   latter   is   divided    internally  or 


proportional    to    the  two 


eX'ternally    into    segments 
adjacent  sides. 

Let  AP  and  AP^  bi- 
sect the  interior  and 
exterior  angles  of  the 
triangle  ABC,  and  meet 
the  opposite  sides  in 
P  and  P\ 

First,  to  prove  that 

BA  :  AC  =  BP  :  PC. 

Draw  CD  parallel  to  the  bisector  PA,  meeting  the  pro- 
longation of  the  side  J5^  in  the  point  D. 

Outline.  By  hypothesis  and  the  properties  of  parallels, 
prove  the  angles  ADC  and  ACD  equal;  and  AD  equal  to  AC. 
Then  use  theorem  1. 


Next,  to  prove  that 

BA:  AC 


BP' :  P'C. 


Draw  CD'  parallel  to  the  bisector  p'a,  meeting  the  side 
BA  in  the  point  D'. 

Outline.     Prove  AD'  equal  to  ^C;  and  use  theorem  1. 

Converse, 

54.  Cor.  I.  If  the  base  of  a  triangle  is  divided  internally 
or  externally  in  the  ratio  of  the  sides,  the  line  drawn  from  the 
point  of  division  to  the  vertex  bisects  the  interior  or  exterior 
vertical  angle. 

For  there  is  only  one  point  P  in  which  BC  can  be  divided 
internally  so  that  BP :  PC  shall  be  equal  to  a  given  ratio ; 
and  only  one  point  P'  in  which  BC  can  be  divided  externally 
so  that  BP' :  P'c  shall  be  equal  to  a  given  ratio  (12,  13). 


294  PLANE  GEOMETRY  —  BOOK  V 

HARMONIC    DIVISION 

55.  Definition.  When  a  line  is  divided  internally  and 
externally  into  segments  having  equal  ratios,  the  line  is 
said  to  be  divided  harmonically. 

56.  Cor.  2.  In  a  triangle  the  bisectors  of  an  interior  and 
its  adjacent  exterior  angle  divide  the  opposite  side  harmonically. 

57.  Cor.  3.  The  hypotenuse  of  a  right  triangle  is  cut 
harmonically  by  any  two  lines  through  the  vertex  of  the  right 
angUy  making  equal  angles  with  one  of  the  sides. 

Harmonic  conjugates.  The  following  corollaries  are  im- 
mediate inferences  from  the  above  definition  and  from 
certain  previous  propositions. 

58.  Cor.  I.  If  a  line  LL'  is  divided  harmonically  at  the 
points  M  and  M\  then  the  line  MM'  is  divided  harmonically  at 
the  points  L  and  L'. 


L  M  L'  M' 

Outline.  Given  LM  :  ML'  =  LM' :  L'm' ;  prove  by  recipro- 
cation and  alternation  that  L'm'  :  ML'  =  LM'  :  LM  (IV.  47,  51). 

59.  Definition.  When  the  line  LL '  is  divided  harmonically 
at  the  points  M  and  m',  then  the  four  points  i,  if,  L',  and  M' 
are  said  to  form  a  harmonic  range.  The  points  M  and  M' 
are  said  to  be  harjnonic  conjugates  with  regard  to  the 
points  L  and  L';  and  the  points  L  and  L'  are  said  to  be 
harmonic  conjugates  with  regard  to  the  points  M  and  m'. 

60.  Cor.  2.  Given  any  three  collinear  points  L,  M,  L'  ;  to 
find  the  harmonic  conjugate  of  M  with  regard  to  L  and  L'. 

[Divide  LV  externally  in  the  ratio  LM :  ML'  (13).] 

61 .  Cor.  3.  A  point  has  only  one  harmonic  conjugate  with 
regard  to  tico  other  collinear  points. 


SIMILAR  POLYGONS  295 

SIMILAR  POLYGONS 

This  section,  which  treats  of  the  construction  and  proper- 
ties of  similar  polygons,  is  based  on  the  properties  of  similar 
triangles  established  in  the  preceding  articles. 

Construction  of  Similar  Polygons 

62.  Problem  7.  To  construct  a  polygon  similar  to 
a  given  polygon  and  such  that  the  ratio  of  similitude 
shall  be  equal  to  a  given  ratio. 

Let    ABCD     be 

the  given  polygon,  /\^  ,.--'1  L 

and  Z :  iJf  the  given  /   ^y^\       '7"^^'     1        M 

ratio.  /     ^.---Vs'         i        \ 

To    construct    a    ^^^^^"^"^-^  "\    \       \ 
Similar  polygon  ^^---^^^^^  ~^n^     1 

A'b'c'd'  such  that  ^^^""^^^^z> 

the    ratio    of    two 

corresponding  sides  AB  :  A'B^  may  be  equal  to  the  given 
ratio  L  :  M.     (Similar  in  construction  and  proof  to  41.) 

63.  Note.  As  in  42,  the  new  polygon  is  called  an  enlarge- 
ment or  reduction  of  the  given  one  according  as  the  consequent 
of  the  given  ratio  is  greater  or  less  than  the  antecedent. 

64.  Cor.  If  a  polygon  P  is  similar  to  a  polygon  Q,  and  if 
Q  is  similar  to  a  third  polygon  R,  then  P  is  similar  to  R  ;  and 
the  ratio  of  similitude  of  P  to  R  is  equal  to  the  ratio  compounded 
of  the  two  ratios  of  similitude  of  P  to  Q,  and  Q  to  R. 

From  the  hypothesis,  the  three  polygons  are  mutually 
equiangular.  Let  L,  L',  and  L"  be  any  corresponding  sides. 
Then  the  ratio  L  :  L"  is  by  definition  compounded  of  the 
ratios  L  :  L'  and  L' :  L"y  that  is,  of  the  two  ratios  of  simili- 
tude. Hence  P  and  R  have  each  pair  of  corresponding  sides 
in  the  same  ratio,  and  are  therefore  similar. 

Ex.  If  a  given  polygon  is  first  enlarged  in  a  ratio  of  similitude  equal 
to  2  :  5,  and  the  result  reduced  as  3  : 1,  what  is  the  whole  alteration  ? 

MOM.  ELEM.  GEOM. — 20 


296  PLANE  GEOMETRY— BOOK  V 

65.  Definition.  Two  broken  lines  are  said  to  be  similar  if 
the  segments  of  one  taken  in  order  as  antecedents,  form  a 
series  of  equal  ratios  with  the  segments  of  the  other  taken 
in  order  as  consequents,  and  if  the  angle  between  two  adja- 
cent segments  equals  the  angle  between  the  corresponding 
segments.  The  ratio  of  two  corresponding  segments  is  called 
the  ratio  of  similitude  of  the  two  similar  broken  lines. 

Ex.  To  construct  a  broken  line  similar  to  a  given  broken  line  and 
such  that  they  shall  have  a  given  ratio  of  similitude. 

Similar  polygon  on  given  line, 

66.  Problem  8.  On  a  given  line  to  construct  a  polygon 
similar  to  a  given  polygon,  and  such  that  the  given  line  shall 
correspond  to  an  assigned  side  of  the  given  polygon. 

Let  ABCD  he  the  given  polygon,  and  L  the  given  line. 


D" f 

\ 


^'f^^-..  ■■...-.\^-  ->B'  /  \ 


L 


J! 

To  construct  a  polygon  on  L  similar  to  ABCD,  and  such 
that  L  shall  correspond  to  the  side  AB. 

Draw  a  line  A'B^  parallel  to  AB  and  equal  to  the  given 
line  L.  Let  the  lines  AA^  and  BB^  meet  at  0.  Draw  5'c' 
parallel  to  BC,  meeting  DC  at  C'.  Draw  ^'Z>'  parallel  to  ADj 
meeting  OD  at  D' ;  and  join  CfD\ 

Outline.  Prove  OC' :  DC  =  OD' :  CD  ;  and  hence  prove  C'd' 
parallel  to  CD.  Show  that  ABCD  and  A'b'c'd'  are  mutually 
equiangular.     Also  prove  that 

A'B' :  AB  =  B'C'  :  BC  —  C'D'  :  CD  =  D'A'  :  DA. 

Then  transfer  A'b'c'd'  to  the  position  a"b"c"d"  (I.  199). 


SIMILAR  POLYGONS  297 

Properties  of  Similar  Polygons 

Transference  into  parallelism, 

67.  Theorem  14.  If  two  polygons  are  similar,  one 
of  them  can  always  he  so  transferred  that  the  corre- 
sponding sides  shall  he  parallel. 

If  the  polygons  are  directly  similar,  turn  one  of  them  about 
a  vertex  by  the  rotation  construction  (I.  202)  until  a  side 
becomes  parallel  to  its  corresponding  side. 

Each  side  of  one  is  then  parallel  to  the  corresponding  side 
of  the  other  (I.  212). 

If  the  polygons  are  obversely  similar,  obvert  one  of  them 
with  regard  to  a  convenient  axis  (I.  227).  The  obverse  is 
then  directly  similar  to  the  given  one.     Rotate  as  before. 

68.  Note.  As  one  line  can  be  rotated  into  parallelism 
with  another  in  either  of  two  ways,  there  are  two  cases  to  be 
distinguished,  as  in  the  following  definition. 

69.  Definition.  When  two  similar  polygons  are  placed 
so  that  corresponding  sides  are  parallel,  the  polygons  are 
said  to  be  plojced  in  parallelism.  When  the  parallel  sides 
are  at  the  same  side  of  the  line  joining  corresponding 
extremities,  the  polygons  are  said  to  be  similarly  placed. 
When  the  parallel  sides  are  at  opposite  sides  of  the  line 
joining  corresponding  extremities,  the  polygons  are  said  to 
be  oppositely  placed. 

Concurrence  of  certain  lines, 

70.  Theorem  15.  If  two  similar  polygons  are  placed 
in  parallelism,  then  the  lines  joining  corresponding 
vertices  are  concurrent ;  except  when  tlie  polygons  are 
equal  and  similarly  plojced,  in  which  case  the  lines 
joining  corresponding  vertices  are  parallel. 

Let  the  two  similar  polygons  ABCD,  A'b'c'd'  be  so  placed 
that  the  sides  AB,  BC,  CD,  DA  are  parallel  respectively  to 
the  homologous  sides  A'b',  b'c',   C'd',  d'A'j    and  let  the 


298 


PLANE  GEOMETRY  —  BOOK   V 


polygons  be  similarly  placed  in  the   left-hand   figure,  and 
oppositely  placed  in  the  right-hand  figure. 


To  prove  that  the  four  lines  A  A',  bb\  ccfj  and  DD'  are  con- 
current in  all  cases,  except  when  the  polygons  are  equal 
and  similarly  placed. 

In  the  left-hand  figure,  if  the  polygons  are  equal,  the 
quadrangle  abb' A'  is  a  parallelogram  ;  hence  AA'  is  parallel 
to  BB\     Similarly  BB'  is  parallel  to  C'C';  and  so  on. 

If  the  polygons  in  the  left-hand  figure  are  not  equal,  pro- 
long AA'  to  meet  BB'  in  0. 

The  triangles  ABO  and  A'B^O  are  similar  (38). 

Therefore  AO:  A'0  =  AB:  a'b'. 

That  is  to  say,  the  line  A  A'  is  divided  externally  by  the 
line  BB'  in  the  ratio  of  similitude  of  the  polygons. 

Similarly  the  same  line  AA'  is  divided  externally  by  each 
of  the  lines  CCf  and  dd'  in  the  same  ratio. 

Hence  the  four  lines  intersect  in  the  same  point  (13). 

Next  take  the  right-hand  figure,  in  which  the  similar 
polygons  are  oppositely  placed.  Then  by  similar  reasoning, 
the  line  BB'  cuts  AA'  internally  in  the  ratio  of  similitude. 

Similarly  A  A'  is  cut  internally  in  the  same  ratio  by  CCf 
and  dd'.  Hence  the  lines  A  A',  BB',  CCf,  and  dd'  intersect  in 
the  same  point.  In  this  case  the  proof  holds  even  when  the 
polygons  are  equal. 


SIMILAR  POLYGONS  299 

71.  Definition.  The  point  of  concurrence  of  the  lines 
joining  corresponding  vertices  of  two  similar  polygons 
placed  in  parallelism,  is  called  the  center  of  similitude  of 
the  polygons. 

72.  Cor.  I.  If  a  line  is  drawn  through  the  center  of  simili- 
tude to  meet  two  corresponding  sides  of  the  polygons,  then  the 
segment  intercepted  between  these  sides  is  divided  at  the  center 
of  similitude  {internally  or  externally)  in  the  ratio  of  similitude. 

Ex.  1.  If  two  similar  broken  lines  are  placed  with  corresponding 
segments  parallel,  then  the  lines  joining  corresponding  vertices  are 
either  parallel  or  concurrent. 

Ex.  2.  If  two  similarly  divided  straight  lines  are  placed  parallel, 
then  the  lines  joining  corresponding  points  of  division  are  either 
parallel  or  concurrent. 

Ex.  3.  If  the  vertices  of  a  polygon  are  joined  to  a  given  point, 
and  if  the  joining  lines  are  each  divided  internally  (or  externally)  in 
any  given  ratio,  then  the  lines  joining  the  points  of  division  form  a 
polygon  similar  to  the  given  one  and  similarly  (or  oppositely)  placed. 
(Converse  of  7.2.     Compare  62.) 

If  the  given  (internal)  ratio  is  equal  to  3  :  5,  show  that  the  ratio  of 
similitude  is  equal  to  8  :  5. 

Principle  of  Correspondence 

73.  Definition.  Any  two  points  are  said  to  be  similarly 
■placed  (or  to  correspond)  with  regard  to  any  two  similar 
polygons,  respectively,  if  the  triangles  having  corresponding 
sides  for  bases,  and  the  two  points  for  vertices,  are  similar 
in  pairs. 

Construction  of  corresponding  points. 

74.  Problem  9.  Given  any  two  similar  polygons, 
and  a  certain  point  within,  without,  or  on  the  bound- 
ary of  the  first  polygon;  to  find  a  similarly  placed 
point  with  regard  to  the  second. 

Let  ABCD,  A'b'c'd'  be  the  given  similar  polygons;  and 
let  0  be  the  given  point. 


300  PLANE  GEOMETRY — BOOK  V 

To  find  a  point  O'  such  that  0  and  0'  shall  be  similarly 
placed  with  regard  to  the  similar  polygons. 


Join  0  to  two  adjacent  vertices  A  and  B.  Draw  A'O', 
making  the  angle  b'a'O'  equal  to  BAG.  Also  draw  B'O'j 
making  the  angle  A'b'O'  equal  to  ABO.  Let  these  lines 
intersect  in  o'. 

Then  O'  corresponds  to  0. 

Since  the  triangles  GAB  and  G'a'b'  are  similar,  hence  the 
ratio  BG:  b'O'  equals  the  ratio  of  similitude  of  the  polygons, 
and  therefore  equals  the  ratio  BC:  B'C'. 

Also  the  angles  GBC  and  G'b'C'  are  equal,  being  the 
differences  of  angles  that  are  respectively  equal. 

Hence  the  triangles  GBC  and  G'B'C'  are  similar. 

Similarly  the  other  triangles  whose  vertices  are  at  0  and 
0'  are  similar  in  pairs. 

Therefore  0'  corresponds  to  G,  by  definition. 

Discussion.  Prove  that  no  other  point  but  0'  can  corre- 
spond to  G. 

Show  that  when  0  comes  nearer  and  nearer  to  one  of  the 
sides,  then  0'  comes  nearer  and  nearer  to  the  corresponding 
side. 

If  0  coincides  with  one  vertex,  prove  that  O'  then  coin- 
cides with  the  corresponding  vertex. 

Show  that  a  similar  construction  applies  when  the  given 
point  is  within  the  first  given  polygon. 


SIMILAR  POLYGONS  301 


CORRESPONDING   LINES 


76.  Definition.  The  lines  joining  two  corresponding  points 
(with  regard  to  two  similar  polygons)  are  called  corre- 
sponding lines. 

From  the  definitions  of  corresponding  points  and  lines, 
and  from  the  properties  of  similar  triangles,  the  following 
corollaries  are  easily  derived. 

76.  Cor.  I.  In  two  similar  polygons,  the  lines  joining  two  cor- 
responding points  to  two  corresponding  vertices  are  in  the  ratio 
of  similitude.     (Use  73,  and  a  property  of  similar  triangles.) 

77.  Cor.  2.  In  tivo  similar  polygons,  any  two  correspond- 
ing lines  are  in  the  ratio  of  similitude. 

Outline.  Let  ABCD  and  A'b'c'd'  be  the  polygons.  Let 
the  point  P  correspond  to  P',  and  Q  to  Q'. 

To  prove  that  the  ratio  PQ :  P'Q'  is  equal  to  the  ratio  of 
similitude. 

The  angles  PAQ  and  p'a'q'  are  equal,  being  differences  of 
equal  angles.  Also  AP  :  A'P'  =  AQ:  A'Q',  each  ratio  being 
equal  to  the  ratio  of  similitude.     Draw  conclusion. 

78.  Cor.  3.  In  two  similar  polygons,  two  triangles,  whose 
respective  vertices  are  at  corresponding  points,  are  similar. 

Use  77,  and  conditions  of  similarity. 

79.  Cor.  4.  If  two  polygons  are  similar,  then  two  other 
polygons,  whose  respective  vertices  taken  in  order  are  at  cor- 
responding points,  are  similar. 

80.  Cor.  5.  If  two  similar  polygons  are  transferred  so  as 
to  he  similarly  placed,  then  any  two  corresponding  lines  become 
parallel. 

Show  that  lines  joining  two  corresponding  points  to  two  correspond- 
ing vertices  become  parallel,  and  apply  76  and  78. 


302 


PLANE  GEOMETRY — BOOK  V 


81.  Ck)r.  6.  If  three  points  are  on  a  straight  line,  then  their 
corresponding  points  are  on  a  straight  line  (80). 

82.  Cor.  7.  If  two  lines  correspond  respectively  to  two 
other  lines  ivith  regard  to  two  similar  j^olygons,  then  the  inter- 
section of  the  first  two  lines  corresponds  to  the  intersection  of 
the  other  two. 

Ex.  1.  In  two  similar  triangles  the  feet  of  the  perpendiculars 
drawn  from  two  corresponding  points  to  the  opposite  sides  are  cor- 
responding points ;  and  these  perpendiculars  are  in  the  ratio  of 
similitude. 

Ex.  2.  In  two  similar  triangles  the  ratio  of  the  radii  of  their  in- 
scribed circles  is  equal  to  the  ratio  of  similitude  ;  so  is  the  ratio  of  the 
radii  of  their  circumscribed  circles. 

Ex.  3.  Given  two  similar  polygons,  to  find  a  point  such  that  it 
coincides  with  its  corresponding  point. 

Analysis.  Suppose  that  P  is  the  required  point.  Let  the  two  cor- 
responding sides  AB  and  A' B'  meet  in  0.  The  triangles  ABP  and 
A'B'P  are  equiangular.  Prove  that  the  quadrangles  OPAA'  and 
OPBB'  are  each  circumscriptible  (III.  62).  Show  that  we  can  deter- 
mine P  by  the  intersection  of  the  circles  described  about  OAA'  and 
OBB' ;  and  prove  that  this  point  is  self-correspondent. 

Note.  This  point  may  be  called  the  center  of  similitude  of  the  two 
similar  polygons.  Show  that  the  center  of  similitude  of  two  similar 
and  similarly  (or  oppositely)  situated  polygons  is  a  special  case  of  this. 

Similar  partition  of  similar  polygons, 

83.  Problem  10.  To  divide  two  similar  polygons 
into  triangles  similar  in  pairs,  and  having  corre- 
sponding points  for  corresponding  vertices. 


SIMILAR  POLYGONS  303 

Outline.  Take  any  point  within  one  of  the  polygons, 
and  find  its  correspondent  (74).  Join  these  two  points  with 
the  vertices  of  the  respective  polygons.  The  triangles  so 
formed  are  similar  in  pairs,  and  have  the  same  ratio  of 
similitude  as  the  polygons  have. 

Note.  The  first  polygon  may  be  dissected  into  triangles 
in  an  arbitrary  manner,  by  taking  as  vertices  any  number  of 
points  inside  the  polygon.  The  second  polygon  can  then  be 
divided  into  similar  triangles  by  joining  the  respective  cor- 
responding points  inside  that  polygon. 

84.  Cor.  When  two  similar  polygons  are  divided  as  in  83, 
then  two  similarly  placed  points  in  a  pair  of  corresponding  tri- 
angles are  also  similarly  placed  with  regard  to  the  polygons. 

POLYGONS    INSCRIBED    IN    POLYGONS 

85.  Definition.  One  polygon  is  said  to  be  inscribed  in 
another  if  each  vertex  of  the  one  is  on  a  side  of  the  other. 

Similar  polygons  in  similar  polygons, 

86.  Problem  11.  Given  two  similar  polygons  and 
given  any  polygon  inscribed  in  the  first;  to  inscribe 
a  similar  polygon  in  the  second. 

Outline.  Let  P  and  P'  be  two  similar  polygons.  Let  a 
polygon  Q  be  inscribed  in  P.     To  inscribe  a  similar  one  in  P'. 

Let  the  vertices  of  Q  hQ  A,  B,  C,  ... ,  all  situated  on  the 
sides  of  P.  Find  their  correspondents  A',  B',  C',  ... ,  all 
situated  on  the  sides  of  P'  (74).  The  inscribed  polygon 
A'b'c'  ...  is  similar  to  the  inscribed  polygon  ABC  ...  (79). 

87.  Cor.  I.  In  a  given  triangle  (P)  to  inscribe  a  tnangle 
similar  to  a  given  triangle  (Q). 

Outline.  Through  the  vertices  of  Q  draw  lines  respec- 
tively parallel  to  the  sides  of  P,  thus  forming  a  triangle 
similar  to  P,  and  having  Q  as  an  inscribed  triangle.  Then, 
by  means  of  86,  inscribe  in  P  a  triangle  similar  to  Q. 


304  PLANE  GEOMETRY  —  BOOK  V 

88.  Cor.  2.  In  a  given  triangle  (ABC)  to  inscribe  a  parallelo- 
gram similar  to  a  given  x>arallelogram  (LMNP) . 

Outline.  Transfer  LMNP  so  that  NP  may  be  parallel  to 
BC.  Through  L  and  M  draw  parallels  to  AB  and  AC,  thus 
forming  a  triangle  similar  to  ABC,  and  having  LMNP  as  an 
inscribed  parallelogram.  Then,  by  means  of  86,  inscribe  in 
ABC  a  parallelogram  similar  to  LMNP. 

Ex.  1.     In  a  given  triangle  to  inscribe  a  square. 

Ex.  2.    In  a  triangle  to  inscribe  a  rectangle  similar  to  a  given  one. 

RATIO  OF  SURFACES  OF  POLYGONS 

89.  All  the  ratios  hitherto  considered  have  been  ratios  of 
segments  of  lines.  It  will  now  be  shown  how  to  compare 
the  surfaces  of  polygons.  The  ratio  of  the  surfaces  of  two 
polygons  will  be  called  the  ratio  of  the  polygons.  We 
begin  with  the  polygons  that  are  most  easily  compared, 
namely,  two  rectangles  of  equal  altitudes,  and  thence  ad- 
vance, step  by  step,  to  the  comparison  of  polygons  in  general. 

90.  Theorem  16.  If  two  rectangles  have  equal  alti- 
tudes, then  the  ra^io  of  the  rectangles  is  equal  to  the 
ratio  of  their  bases. 

Let  the  rectangles  OJBC,  O'a'b'c'  have  the  bases  OA,  O'a', 

and  the  equal  altitudes  AB,  a'b'.      Let  the  rectangles  be 

denoted  by  R,  R',  and  their  bases 

,      ,     ,,  C'    B'     B'o     p., 

by    h,   b.  r r -r-<-— I 

To  prove     R:  R'  =  b:b'. 

Prolong  OA  and  lay  off  consecu- 
tive segments  equal  to  OA.  Mark  J 
the  points  of  division  with  the 
symbols  A2,  A^  ... ;  and  draw 
perpendiculars  through  these 
points  to  meet  the  prolongation 
of  CB.     Prolong  O'A',  and  make  a  similar  construction. 


i    i    ! 

cr-A'-AVAV"^' 


SURFACE  RATIOS  305 

The  segment  OA^  is  equal  to  mb ;  and  the  rectangle  stand- 
ing on  it  is  equal  to  mR. 

The  segment  o'A'^  is  equal  to  nb' ;  and  the  rectangle  stand- 
ing on  it  is  equal  to  nR'. 

Since  the  altitudes  are  equal,  the  pairs  of  magnitudes 

rect.  OB^,  rect.  OB'^ 
and  base  OA^,  base  0A'„ 

are  in  the  same  order  of  size  (II.  22,  III.  50), 
i.e.,  the  pairs  of  multiples  ^ 

mRj  nR' 
and  mb,  nb' 

are  in  the  same  order  of  size,  whatever  m  and  n  are ;  there- 
fore the  scale  of  relation  of  R  and  R'  is  everywhere  similar 
to  the  scale  of  relation  of  b  and  b' ;  hence,  by  definition  of 
equal  ratios,  R-  r'  =  b  -b' 

Parallelograms  of  equal  altitudes. 

91.  Cor.  I.  Two  parallelograms  of  equal  altitudes  have 
a  ratio  equal  to  the  ratio  of  their  bases. 

Triangles  of  equal  altitudes. 

92.  Cor.  2.  Two  triangles  of  equal  altitudes  have  a  ratio 
equal  to  the  ratio  of  their  bases. 

Cor.  3.  Two  parallelograms  or  triangles  of  equal  bases 
have  a  ratio  equal  to  the  ratio  of  their  altitudes. 

Ex.  1.  Perpendiculars  are  drawn  from  any  point  within  an  equi- 
lateral triangle  on  the  three  sides  :  show  that  their  sum  is  equal  to  the 
altitude  of  the  triangle  (IV.  59,  33). 

Ex.  2.  A  quadrangle  is  divided  by  its  diagonals  Into  four  triangles 
that  form  a  proportion. 

Ex.  3.  If  two  triangles  have  their  bases  in  the  same  straight  line, 
and  their  vertices  on  the  same  line  parallel  to  the  bases,  then  any 
other  parallel,  cutting  the  sides,  cuts  off  two  triangles  that  form  a  pro- 
portion with  the  given  triangles. 


306  PLANE  GEOMETRY  —  BOOK  V 

Relation  among  four  propoHional  lines, 

93.  Theorem  17.  //  four  lines  form  a  proportion, 
then  the  rectangle  of  the  extremes  is  equivalent  to 
the  rectangle  of  the  means. 

Let  a,  6,  c,  d  be  four  lines  such  that        a:b  =  c:d. 


To  prove  that  the  rectangle  of  a  and  d  is  equivalent  to  the 
rectangle  of  b  and  c. 

On  one  of  the  sides  of  any  right  angle  lay  off  OA,  OB  equal 
to  a,  6 ;  and  on  the  other  side  lay  off  OC,  OD  equal  to  c,  d. 
Complete  the  rectangles  AD  and  BC. 

Compare  each  of  these  rectangles  with  the  rectangle  BDj 
which  is  their  common  part. 

Since  rectangles  of  equal  altitudes  have  a  ratio  equal  to 
the  ratio  of  their  bases,  therefore 

rect.  AD  :  rect.  BD  =  OA 
and  rect.  BC :  rect.  BD  =  OC 

now  OA:  OB  =  OC 

therefore,  by  equality  of  ratios, 

rect.  AD  :  rect.  BD  =  rect.  BC :  rect.  BD ; 
since  these  equal  ratios  have  a  common  consequent,  hence  the 
rectangles  AD  and  BC  are  equivalent;  that  is,  the  rectangle 
of  the  extremes  is  equivalent  to  the  rectangle  of  the  means. 

Special  case. 

94.  Cor.  If  three  lines  are  proportional,  the  rectangle  of  the 
extremes  is  equivalent  to  the  square  on  the  mean. 

Ex.  1.     Apply  the  theorem  to  prove  III.  94. 
Ex.  2.     Apply  the  corollary  to  prove  II.  60. 


OB, 

[90 

OD; 

OD, 

[liyp- 

SURFACE  RATIOS  307 

Converse  of  93, 

95.  Theorem 'IS.  If  two  rectangles  are  equivalent, 
the  sides  of  one  will  form  the  extremes,  and  the  sides 
of  the  other  the  means,  of  a  proportion. 

In  figure  of  theorem  17,  if  rectangles  AD  and  BC  are 
equivalent,  they  have  equal  ratios  to  rectangle  BD  (IV. 
25).     Therefore,  etc. 

Converse  of  94- 

96.  Cor.  If  there  are  three  lines  such  that  the  rectangle  of 
the  extremes  is  equivalent  to  the  square  on  the  mean,  then  the 
three  lines  form  a  proportion. 

Extreme  and  Mean  Ratio 

97.  Definition.  If  a  given  line  is  divided  into  two  parts 
such  that  one  of  the  parts  is  a  mean  proportional  between 
the  whole  line  and  the  other  part,  the  line  is  said  to  be 
divided  in  extreme  and  mean  ratio  or  in  medial 
section. 

Application  of  94, 

98.  Problem  12.  To  divide  a  given  line  in  extreme 
and  mean  ratio. 

By  means  of  the  construction  in  II.  89,  divide  the  given 
line  so  that  the  rectangle  of  the  whole  line  and  one  part  is 
equivalent  to  the  square  on  the  other  part. 

The  line  is  then  divided  in  extreme  and  mean  ratio; 
because  the  latter  part  is,  by  96,  a  mean  proportional 
between  the  whole  line  and  the  first  part. 

Note.     This  mode  of  division  is  the  ancient  sectio  aurea  (II.  89). 

Ex.  If  the  radius  of  a  circle  is  divided  in  extreme  and  mean  ratio, 
the  greater  segment  is  equal  to  the  side  of  an  inscribed  regular  deca- 
gon (III.  122). 


308  PLANE  GEOMETRY  —  BOOK   V 

Mutually  Equiangular  Parallelograms 

# 
99.  From  the  comparison  of  two  parallelograms  of  equal 
altitudes,  or  of  equal  bases  (91,  92),  we  can  advance  to  the 
comparison  of  any  two  mutually  equiangular  parallelograms. 
This  is  done  by  introducing  an  intermediate  parallelogram 
having  a  side  in  common  with  each,  and  then  compound- 
ing the  two  successive  ratios.  The  ratio  of  the  two  given 
surfaces  is  thus  expressed  as  a  ratio  compounded  of  two 
line-ratios  by  means  of  the  following  theorem. 

100.  Theorem  19.  If  two  parallelograms  are  mutu- 
ally equiangular,  then  their  ratio  is  equal  to  the  ratio 
compounded  of  the  ratios  of  two  a^acent  sides  of  tlie 
first  to  the  respective  adja^cent  sides  of  the  second. 

Let  ABCD  and  BEFG  be  two  parallelograms  that  have  the 
angles  ABC  and  EBG  equal.  Let  these  parallelograms  be 
noted  by  P  and  R.  ^  en 

To  prove  that  the  ratio  P :  R       /  p  ~7  n" 7 

equals  the  ratio  compounded  of     L _/ J^ 

the  two  ratios  /  i?  / 


AB  :  BG  and  CB  :  BE.  E  Ji 

Place  the  two  parallelograms  so  that  the  sides  AB  and  BG 
are  in  one  line,  and  so  that  the  equal  angles  ABC  and  GBE 
are  vertically  opposite.  Then  the  sides  CB  and  BE  are  in 
one  line  (I.  52). 

Complete  the  parallelogram  BGHC,  and  denote  it  by  Q. 

In  the  set  of  three  magnitudes  P,  Q,  R,  the  ratio  P :  P  is, 
by  definition  (22),  compounded  of  the  successive  ratios  P :  Q 
and  Q  :  R. 

Now  P:  Q=AB:BG,  [91 

and  Q:  R=CB:  BE. 

Therefore  the  ratio  of  the  two  parallelograms  P  and  R  is 
equal  to  the  ratio  compounded  of  the  ratios  of  their  sides. 


SURFACE  RATIOS  309 

Ex.  1.  Given  two  mutually  equiangular  parallelograms,  show  how 
to  convert  their  ratio  into  a  line-ratio,  by  a  construction.     (See  25.) 

Ex.  2.  The  sides  of  one  rectangle  are  to  the  respective  sides  of  an- 
other in  the  two  ratios  3 :  2  and  4:5.  Show  that  the  first  rectangle  is 
to  the  second  as  6  :  5. 

101.  Cor.  I.  The  ratio  compounded  of  two  line-ratios  is 
equal  to  the  ratio  of  the  rectangle  constructed  on  the  antecedents 
to  the  rectangle  of  the  consequents. 

102.  Cor.  2.  If  two  triangles  have  an  angle  of  the  one  equal 
or  supplemental  to  an  angle  of  the  other,  the  ratio  of  the  tri- 
angles is  equal  to  the  ratio  compounded  of  the  two  ratios  of  the 
including  sides  of  the  one  to  the  including  sides  of  the  other. 

[Show  that  the  triangles  are  halves  of  mutually  equiangular 
parallelograms.  ] 

Ex.  If  two  triangles  have  an  angle  of  the  one  equal  to  an  angle  of 
the  other,  and  if  the  including  sides  are  respectively  as  1 :  3  and  1 : 4, 
show  that  the  first  triangle  is  one  twelfth  of  the  second. 

103.  Cor.  3.  The  ratio  of  two  mutually  equiangular  paral- 
lelograms is  equal  to  the  ratio  of  the  two  rectangles  contained 
by  the  adjacent  sides  respectively. 

Ratio  of  similar  parallelograms, 

104.  Theorem  20.  If  two  parallelograms  are  simi- 
lar, then  their  ratio  is  equal  to  the  duplicate  of  their 
ratio  of  similitude. 

For  their  ratio  is  equal  to  the  ratio  compounded  of  the 
ratios  of  two  adjacent  sides  of  one  to  the  corresponding 
sides  of  the  other  (100). 

Now  each  of  these  ratios  is  equal  to  the  ratio  of  simili- 
tude. Hence  the  ratio  of  the  parallelograms  is  equal  to  the 
duplicate  of  the  ratio  of  similitude  (30). 

Ratio  of  similar  triangles, 

105.  Cor.  I.  Two  similar  triangles  have  a  ratio  equal  to 
the  duplicate  of  their  ratio  of  similitude. 


310  PLANE  GEOMETRY — BOOK  V 

106.  Cor.  2.  Two  squares  have  a  ratio  equal  to  the  dupli- 
cate of  the  raiio  of  their  sides. 

107.  Cor.  3.  Two  similar  triangles  have  a  ratio  equal  to 
the  ratio  of  the  squares  on  corresponding  sides. 

Ex.  If  the  ratio  of  similitude  of  two  similar  triangles  is  equal  to 
3:1,  how  often  is  the  less  contained  in  the  greater  ? 

Ratio  op  Similar  Polygons 

108.  Theorem  21.  Two  similar  polygons  have  a 
ratio  equal  to  the  duplicate  of  tJzeir  ratio  of  simili- 
tude.    (Apply  83,  105.) 

108  (a).  Cor.  i.  The  ratio  of  two  similar  polygons  is 
equal  to  the  ratio  of  the  squares  on  corresponding  sides. 

109.  Cor.  2.  A  polygon  is  greater  than^  eqvxd  tOy  or  less 
than  a  similar  polygon^  according  as  a  side  of  the  first  is 
greater  than,  equal  to,  or  less  than  the  corresponding  side  of  the 
second;  and  conversely  (II,  24). 

110.  Cor.  3.  If  three  lines  are  proportional,  then  the  first  is 
to  the  third  as  any  polygon  standing  on  the  first  is  to  the  simi- 
lar and  similarly  situated  polygon  standing  on  the  second. 


Q 


L  M  JV        - 

Use  31  and  108. 

Surface-ratio  converted  into  line-ratio, 

111.  Cor.  4.  7(9  find  two  lines  in  the  ratio  of  tJie  surfaces 
of  two  given  similar  polygons. 

To  two  corresponding  sides  L  and  M  find  a  third  propor- 
tional N.     Then  L  and  N  are  the  required  lines  (110). 

Ex.  1.  To  enlarge  (or  reduce)  the  surface  of  a  given  polygon  P  in 
the  given  ratio  L  :  N.     (Use  C2  and  52,  ex.  2.) 

Ex.  2.     Show  how  to  double  a  given  polygon,  preserving  its  shape. 

Ex.  3.     Construct  a  square  equivalent  to  one  third  of  a  given  one. 


SURFACE  RATIOS 


311 


Sum  of  two  similar  polygons. 

112.  Theorem  22.  In  a  right  triangle,  any  polygon 
standing  on  the  hypotenuse  is  equivalent  to  the 
sum  of  two  similar  and  similarly  situated  polygons 
standing  on  the  sides. 

Let  ABC  be  a  triangle,  right  angled  at  C ;  and  let  P,  Q,  R 
be  similar  and  similarly  situated  polygons  on  AB,  BC,  CA, 
respectively. 


To  prove  that  P  is  equivalent  to  the  sum  of  Q  and  R. 
Draw  CD  perpendicular  to  AB. 

Then,  from  the  similarity  of  the  triangles  ABC  and  CBD 
AB  :  BC  =  BC :  DB.  [46 

Therefore,  by  110, 

P:Q  =  AB:DB. 
Similarly,  P\R  =  AB\AD. 

Hence,  by  reciprocation  and  the  rule  of  addition, 

P :  Q -{-  R  =  AB  :  DB  -\-  AD.         [IV.  51,  59 

Now  AB  equals  DB  +  AD. 

Therefore  P  is  equivalent  to  the  sum  of  Q  and  R  (IV.  33). 

Note.     This  theorem  includes  II.  61  as  a  special  case. 

Ex.    Given  two  similar  polygons,  show  how  to  construct  a  third 
polygon  similar  to  them,  and  equivalent  to  their  sum. 

MCM.  ELEM.  GEOM. 21 


812 


PLANE  GEOMETRY — BOOK  V 


GENERAL  CONSTRUCTION  OF  POLYGONS 

Given  the  shape  and  size. 

113.  Problem  13.     To  construct   a  polygon  similar 
to  one  and  equivalent  to  another  £iven  polygon. 

Let  P  and  Q  be  the  two  given  polygons. 

To  construct  a  polygon 
R  similar  to  P  and  equiva- 
lent to  Q. 

On  AB,  a  side  of  P,  con- 
struct the  rectangle  ABCD 
equivalent  to  P  (II.  71-73). 

On  AD  construct  the  rec- 
tangle ^ZJ-Ei^equivalent  to  Q. 

Find  GH  the  mean  proportional  between  BA  and  AF  {52). 

On  GH  construct  a  polygon  R  similar  to  P  and  such  that 
GH  and  BA  are  corresponding  sides  {(^Q). 

Then  R  is  the  polygon  required. 


Since 

BA:GH=  GHiAF, 

[constr. 

therefore 

BA:AF=  P:R, 

[110 

hence 

BD:AE=P:R. 

[90 

Now  these  equal  ratios  have  equivalent  antecedents  (by 
construction) ;  hence  they  have  equivalent  consequents. 
Thus  R  is  equivalent  to  AE,  and  therefore  to  Q. 
Ex.    Construct  an  equilateral  triangle  equivalent  to  a  given  square. 

Conversion  of  a  polygon-ratio. 

114.   Problem   14.     To  find  two  lines  whose  ratio  is 
equal  to  ths  ratio  of  the  surfaces  of  two  given  polygons. 

Let  P  and  P'  be  the  given  polygons. 

To  find  two  lines  h  and  6'  such  that  P  :  P'  =  6  :  6'. 

As  in  113,  construct  two  rectangles  R  and  b!  respectively 
equivalent  to  P  and  P\  and  having  equal  altitudes  (II.  71-3). 
Let  the  bases  of  the  rectangles  be  denoted  by  6  and  6'. 
Then  P  :  P' =  P:  iJ' =  6  :  6'.  [IV.  25,  V.  90 


SURFACE  RATIOS  313 

Ex.  Given  any  two  ratios,  P :  Q  and  R  :  S,  show  how  to  con- 
vert them  into  line-ratios  having  equal  consequents  ;  and  hence  show 
how  to  determine  which  of  the  two  given  ratios  is  the  greater. 

Addition  of  Eatios 

115.  Definition.  If  two  ratios  have  the  same  consequent, 
then  a  third  ratio  having  the  same  consequent  and  having 
its  antecedent  equivalent  to  the  sum  of  their  antecedents  is 
called  the  sum  of  the  first  two  ratios. 

E.g.,  the  sum  of  the  ratios  A  :  B  and  A^ :  B  \?,  A  -\-  A^ :  B. 

The  sum  oi  any  two  ratios  is  the  ratio  obtained  by  con- 
verting them  so  as  to  have  equivalent  consequents,  and  then 
taking  the  sum  by  the  preceding  definition. 

To  add  two  ratios, 

116.  Pkoblem  15.  To  construct  a  line-ratio  equal 
to  the  sum  of  any  two  given  ratios,  whether  of  lines, 
polygons,  or  whole  numbers. 

Outline.  If  either  ratio  is  not  a  line-ratio,  convert  it  into  a  line- 
ratio  (114  ;  16,  ex.).  Convert  these  two  line-ratios  so  as  to  have  a 
common  consequent  (17).  Then  the  required  ratio  has  the  same  con- 
sequent, and  its  antecedent  is  the  sum  of  the  new  antecedents. 

Ex.  Show  that  the  sum  of  the  two  numerical  ratios  m  :  n  and 
p  :  q  \&  mq  +  np  :  nq. 

Equals  added  to  equals, 

117.  Theorem  23.  If  equal  ratios  are  added  respec- 
tively to  equal  ratios,  then  the  sums  are  equal. 

Outline.  Convert  the  ratios  so  that  those  which  are  to 
be  added  may  have  common  consequents.  Let  the  given 
equalities  then  be  written, 

A:B  =  X:Y, 
A':B  =  X^:Y. 
Then  by  the  rule  of  addition  (IV.  59), 

A-\-A^:B  =  X-\-X':  T. 
Now  these  ratios  are  the  sums  of  the  given  ratios  (def., 
115).     Hence  the  sums  are  equal. 


314  PLANE  GEOMETRY  —  BOOK   V 

Addition  commutcubive. 

118.  Cor.  Tlie  sum  of  two  ratios  is  the  same^  however  they 
are  converted  so  as  to  have  a  coinmon  consequent  and  in  wliai- 
ever  order  they  are  taken. 

Ex.  Frame  a  similar  definition  for  the  difference  of  two  ratios; 
and  show  that  if  equal  ratios  are  subtracted  respectively  from  equal 
ratios,  then  the  differences  are  equal. 

Equals  added  to  unequeUs, 

119.  Theorem  24.  If  one  ratio  is  greater  than  an- 
other, tJien  the  sum  of  the  greater,  and  any  third 
ratio  is  greater  than  the  sum  of  the  less  and  the 
same  third  ratio. 

Convert  all  the  ratios  as  before,  and  apply  IV.  28,  29. 

120.  Cor.  If  two  ratios  are  respectively  greater  than  two 
others,  then  the  sum  of  the  two  greater  ratios  is  greater  than 
the  sum  of  the  two  less  ratios. 

DISTRIBUTIVE    PROPERTY    OF    RATIOS 

Cofnjyounding  a  sutn  ivith  a  third  ratio, 

121.  Theorem  25.  If  the  sum,  of  two  ratios  is  com- 
pounded with  any  third  ratio,  the  result  is  tlie  same 
as  if  the  two  ratios  are  first  separately  compounded 
with  the  third  rajbio,  and  the  results  then  added. 

Convert  the  first  two  ratios  into  line-ratios  with  a  common 
consequent;  let  them  be  ^  :  5  and  A*  -.  B.  Convert  the  third 
ratio  into  any  line-ratio  R :  S. 

To  prove  that  the  ratio  compounded  of  A  -\-  A^  -.B  and 
2? :  S  is  equal  to  the  sum  of  the  ratio  compounded  of  ^  :  ^ 
and  R :  S,  and  the  ratio  compounded  of  .i' :  B  and  R :  S. 

Outline.  Replace  each  compound  ratio  by  the  ratio  of  two 
rectangles  (101)  ;  the  resulting  ratios  have  a  common  conse- 
quent, namely  rect.  {B,  S).  To  the  antecedents  apply  the 
distributive  property  of  rectangles  (II.  41),  i.e.  that  rect. 
(A-^  a\  R)  is  equivalent  to  the  sum  of  rect.  (A,  R)  and  rect. 
(^',*  R).     Draw  conclusion  by  definition  in  115. 


RATIOS  IN   THE  CIRCLE 


315 


RATIOS   IN  THE  CIRCLE 
Arc-ratios  and  Angle-ratios 

Equiradial  arcs, 

122.  Theorem  26.  In  two  equal  circles  or  in  the 
same  circle  the  ratio  of  any  two  central  angles  is 
equal  to  the  ratio  of  their  subtending  arcs. 

Let  AB,  A'b'  be  arcs  of  equal  circles  subtending  the  central 
angles  AOB,  a'o'b'.  Let  the  arcs  be  denoted  by  a,  b,  and 
the  angles  by  a,  13. 


To  prove  that  a:  ft  =  a:b. 

Take  the  arc  ABL  equal  to  m  times  a,  and  arc  A'b'l'  equal 
to  n  times  6,  it  being  understood  that  either  or  both  of  these 
multiples  may  exceed  a  whole  circle,  and  accordingly  that 
the  corresponding  central  angles  may  exceed  a  perigon. 

The  arc  ABL  is  made  up  of  m  parts  each  equal  to  AB,  and 
each  subtending  an  angle  a,  hence  the  arc  ma  subtends  an 
angle  equal  to  m«.  Similarly  the  arc  nb  subtends  an  angle 
equal  to  n(S. 

Hence,  from  the  order-theorem  of  arcs  and  central  angles, 
the  two  pairs  of  multiples 

ma,  nb 
and  ma,  nfi 

are  in  like  order  of  size  whatever  m  and  n  are  (III.  35,  51). 
Therefore,  by  definition  of  equal  ratios, 

a:  ^  =  a:b. 


316  PLANE  GEOMETRY — BOOK  V 

123.  Cor.  In  two  equal  circles,  or  in  the  same  circle,  the 
ratio  of  any  two  sectors  is  equal  to  the  ratio  of  their  arcs,  and 
also  equal  to  the  ratio  of  their  angles. 

Inscribed  Polygons 
Rectangle  of  sides  of  inscribed  triangle. 

124.  Theorem  27.  The  rectangle  of  two  sides  of  a 
triangle  is  equivalent  to  the  rectangle  of  two  lines 
drawn  from  the  vertex,  making  equal  angles  with 
these  sides  respectively,  one  of  the  lines  being  ter- 
minated by  the  base,  and  tJie  other  by  tJie  arc  of 
the  circumscribed  circle  below  the  ba^e. 

Let  ABEC  be  the  circumscribed  circle  of  the  triangle  ABC. 
Let  the  lines  AD,  AE  be  drawn  through  the  vertex  A  making 
with  the  sides  AB,  AC  the  equal  angles  BAD,  CAE  respec- 
tively. Let  AD  be  terminated  by  the  base,  and  AE  by  the 
arc  BC  below  the  base. 


To  prove  that  the  rectangle  of  AB  and  AC  is  equivalent  to 
the  rectangle  of  AD  and  AE. 

[Draw  EC.  Prove  the  triangles  ACE  and  ADB  mutually 
equiangular  (III.  54) ;  and  apply  38,  93.] 

125.  Cor.  I.  Tlie  rectangle  of  two  sides  of  a  triangle  is 
equivalent  to  the  rectangle  contained  by  the  diameter  of  the 
circumscribed  circle  and  the  altitude  drawn  to  the  base. 

126.  Cor.  2.  The  rectangle  contained  by  two  sides  of  a 
triangle  is  equivalent  to  the  rectangle  contained  by  two  lines. 


RATIOS  IN   THE  CIRCLE  317 

one  of  which  bisects  the  vertical  angle  and  is  terminated  by 
the  circumscribed  circle,  and  the  other  is  the  portion  of  that 
bisector  intercepted  between  the  vertex  and  the  base. 

127.  Cor.  3.  Hie  rectangle  of  two  sides  of  a  triangle  is 
equivalent  to  the  square  of  the  line  that  bisects  the  vertical 
angle  and  is  terminated  by  the  base,  together  with  the  rectangle 
of  the  segments  of  the  base  made  by  that  bisector. 

Sides  of  inscribed  quadrangle, 

128.  Theorem  28.  If  a  quadrangle  is  inscribed 
in  a  circle,  the  rectangle  of  its  diagonals  is  equiva- 
lent to  the  sum  of  the  two  rectangles  contained 
respectively  by  each  pair  of  opposite  sides. 

Let  ABCD  be  the  inscribed  quadrangle.    ' 


To  prove  that  the  rectangle  oi  AC  and  BD  is  equivalent 
to  the  sum  of  the  rectangle  of  AB  and  CD  and  the  rectangle 
of  AD  and  BC. 

Draw  AE,  making  the  angle  BAE  equal  to  the  angle  CAD. 

Outline.  Prove  the  triangles  ABE  and  ACD  similar;  and 
rectangle  \_AB,  CD']  equivalent  to  rectangle  \_AC,  BE].  Also 
prove  rectangle  \_AD,  BC]  equivalent  to  rectangle  \_AC,  DE]. 
Add,  and  apply  II.  39. 

Note.  This  result  is  called  Ptolemy's  theorem,  after  Claudius 
Ptolemaeus  of  Alexandria,  one  of  the  chief  geometers  among  the  later 
Greeks  (died  165  a.d.). 

Ex.  If  the  three  vertices  of  an  equilateral  triangle  are  joined  to  a 
point  on  the  circumscribed  circle,  then  one  of  the  joining  lines  is  equal 
to  the  sum  of  the  other  two. 


318  PLANE  GEOMETRY  —  BOOK   V 

CIRCUMSCKIPTIBLE    POLYGONS 

129.  Theorem  29.  If  a  polygon  is  circumscriptible, 
then  any  similar  polygon  is  also  circunvscriptihle. 

Let  the  similar  polygons  be  ABC  •••,  A'b'c'  •••,  and  let  the 
former  be  circumscriptible. 

To  prove  that  the  latter  is  also  circumscriptible. 

Find  0,  the  center  of  the  circumscribing  circle  of  the 
former;  and  find  its  corresponding  point  O'  in  the  other 
polygon  (74). 

Then,  by  similar  triangles, 

OA  :  O'A'  =  OB  :  O'B'  =  O'C:  O'Cf  =  •... 

But  OA  =  OB=  OC=z  •". 

Hence  O'A'  =  o'B'  =  o'c'  =  •••. 

Therefore,  the  circle  described  with  O'  as  center  and  O'A' 
as  radius  passes  through  all  the  vertices  of  the  polygon 
a'b'c'  "'.     This  polygon  is  therefore  circumscriptible. 

130.  Cor.  I.  If  two  similar  polygons  are  circumscriptible, 
the  centers  of  the  circumscribed  circles  are  corresponding  points, 
and  the  ratio  of  the  radii  is  equal  to  the  ratio  of  similitude  of 
the  polygons  (76). 

131.  Cor.  2.  If  a  polygon  is  such  that  a  circle  can  be  in- 
scribed, then  any  similar  polygon  has  the  same  property,  and 
the  centers  of  the  inscribed  circles  are  corresponding  points. 

132.  Theorem  30.  The  perimeters  of  any  two  regu- 
lar ii'gons  have  a  ratio  equal  to  the  ratio  of  the  radii 
of  their  eircumscrihed  circles,  and  also  equal  to  the 
ratio  of  the  raxlii  of  their  inscribed  circles. 

Use  130,  131,  and  45,  ex.  2. 

133.  Cor.  Tlie  ratio  of  the  surfaces  of  two  similar  poly- 
gons is  equM  to  the  ratio  of  the  squares  on  the  radii  of  their 
circumscribed  circles. 


RATIOS  IN   THE  CIRCLE  319 

134.  Problem  16.  To  inscribe  in  a  given  circle  a 
polygon  similar  to  a  given  circumscriptible  polygon. 

Outline.  Let  ABCD  be  the  given  polygon,  0  the  center  of 
its  circumscribed  circle,  and  O'  the  center  of  the  given  circle. 

Draw  radii  O'^',  0'B\  O'c',  O'd'  parallel  to  the  radii  OA, 
OB,  OC,  OD. 

The  inscribed  polygon  A'b'C'b'  is  similar  to  ABCD. 

Cor.  In  a  given  circle  to  inscribe  a  rectangle  similar  to  a 
given  rectangle. 

Ex.  1.  In  a  given  semicircle  inscribe  a  rectangle  similar  to  a  given 
rectangle  ABCD. 

[Place  the  rectangle  so  that  any  side  AB  is  parallel  to  the  diameter 
of  the  semicircle.  Bisect  AB  in  0.  Draw  OC,  OD.  In  the  semicircle 
draw  radii  O'C,  O'D'  parallel  to  OC,  OD.  Complete  the  rectangle 
A'B'G'D'.  Prove  it  similar  to  ABCD.  Obtain  another  solution  by 
placing  the  side  BC  parallel  to  the  diameter.  When  are  the  two 
solutions  identical  ?] 

Ex.  2.  In  a  given  quadrant  inscribe  a  rectangle  similar  to  a  given 
rectangle.     How  many  solutions  are  possible  ? 

Ex.  3.  Given  the  ratio  of  two  lines,  and  their  mean  proportional, 
find  them.  [Take  two  lines  in  the  given  ratio,  find  their  mean  propor- 
tional, enlarge  or  reduce  the  figure  to  suit  the  condition  (42).] 

135.  Regular  n-gons  and  2  n-gons.  When  an  inscribed 
regular  n-gon  is  given,  it  has  been  shown  that  if  the  arcs 
are  bisected,  then  the  chords  of  the  half  arcs  form  an  in- 
scribed regular  2  n-gon,  and  that  the  tangents  at  the  vertices 
of  the  latter  form  a  circumscribed  regular  2  n-gon. 

It  has  also  been  seen  that  the  perimeter  of  the  inscribed 
n-gon  is  less  than  the  perimeter  of  the  inscribed  2  7i-gon, 
that  the  latter  perimeter  is  less  than  the  perimeter  of  the 
circumscribed  2  n-gon,  and  that  this  is  in  turn  less  than  the 
perimeter  of  the  circumscribed  7i-gon.  It  is  important,  for 
reasons  which  will  appear  later,  to  obtain  more  definite  rela- 
tions among  these  four  perimeters.  Theorem  31  establishes 
a  simple  relation  among  the  first  three,  and  theorem  32 
gives  a  relation  among  the  first,  third,  and  fourth. 


320  PLANE  GEOMETRY — BOOK  V 

136.  Theorem  31.  If  in  a  circle  are  inscribed  a 
regular  n-gon  and  a  regular  2n-gon,  and  if  about 
the  circle  is  circumscribed  a  regular  2  n-gon,  then 
the  perimeters  of  these  three  polygons  form  a  pro- 
portion. 

Let  AB  be  the  side  of  an  inscribed  regular  n-gon,  and  C 
the  mid-point  of  the  arc  AB-^  then  the  lines  ^C  and  5  Care 
sides  of  an  inscribed  reg- 
ular 2  w-gon,  and  the  two  ^       J^  __        E 

tangents  at  A  and  B  inter- 
cept  on   the   tangent  at 

C  a  line  DE  equal  to  the  _ 

side  of   a  circumscribed 

regular  2  7i-gon.     The  perimeters  of  these  three  polygons 
are,  respectively,  ^.^^^  2n-AC,2n.DE 

To  prove      n-  AB:2n'  AC=2n'  AC:2n'  DE. 

The  angles  ACD  and  ABC  are  equal.  Therefore  the  isos- 
celes triangles  ACD  and  ^fiCare  equiangular  (III.  76). 

Hence  AB  :  AC  =  AC :  DC. 

Now  n:2n  =  2n:4ri. 

Hence,  by  compounding  equal  ratios, 

n  '  AB  .2  n  '  AC  =2  n  '  AC :  4:71 '  DC.  [23,  26 

But  4  n'DC—2n'DE'^ 

therefore        ri'  AB'.2n  -  AC  =  2n'  AC:2n'  DE. 

Note.  If  the  perimeters  of  the  inscribed  and  circum- 
scribed regular  7i-gons  are  denoted  by  p,„  P„,  then  this 
result  may  be  written  in  the  form,  p,, :  P^^  =  Po„  :  po^. 

137.  Theorem  32.  If  in  a  circle  a  regular  n-gon  is 
inscribed,  and  if  about  the  circle  are  circumscribed 
a  regular  n-gon  and  a  regular  2  n-gon,  then  the  sum 
of  the  perimeters  of  the  first  and  second,  of  these 
polygons  is  to  ths  perimeter  of  the  first  as  the  perim- 
eter of  the  second  is  to  the  semiperimeter  of  the  third. 


RATIOS  IN  THE  CIRCLE 


321 


Let  AB  be  the  side  of  a  regular  inscribed  n-gon,  and  draw 
the  lines  as  in  the  preceding  figure.  Join  the  center  0  to 
A,  B,  D,  E.  Prolong  OA  and  OB  to  meet  DE  prolonged  in  F 
and  G;  then  FG  is  evidently  a  side  of  a  regular  circum- 
scribed n-gon.  The  perimeters  of  the  inscribed  n-gon,  the 
circumscribed  n-gon,  and  the  circumscribed  2  n-gon  are, 
respectively,  n  -  AB,  n  -  FG,  2  n  •  DE. 

The  semiperimeter  of  the  latter  polygon  is  n  •  DE, 
F  D         a         E  G 


^x 


AB  =  71'  FG  :  n  '  DE. 


To  prove  n  •  AB  ■\-  n  •  FG  :  n 
By  similar  triangles, 

FG:  AB  =  OF:  OA, 
=  OF :  OC. 
By  equality  of  right  triangles,  OD  bisects  the  angle  COA. 
Hence  OF:  OC  =  FD  :  DC.  [53 

Therefore,  by  equality  of  ratios, 

FG:  AB  =  FD:  DC. 
Hence,  by  composition, 

FG  -}-  AB:  AB  =  FD  +  DC:  DC,  [IV.  48 

=  EC :  DC, 
=  FG :  DE. 
Therefore,  taking  ?ith  multiples, 

n  '  FG  -{-  n  •  AB  :  n  '  AB  =  n  '  FG  :  n  '  DE. 

Note.     With  the  same  notation  as  in  note  to  136,  this 
result  may  be  written  in  the  form, 


322 


PLANE  GEOMETRY  —  BOOK  V 


LOCUS   PROBLEMS 

138.  Problem  17.  To  find  tJie  locus  of  a  point  such 
that  the  perpendiculars  from  it  to  two  given  lines 
shall  have  a  given  ratio. 

Let  OR  and  OS  be  the  given  lines.  Let  P  be  a  point  such 
that  the  perpendiculars  PM  and  PN  have  a  given  ratio  A  :  B. 


To  find  the  locus  of  the  point  P. 

As  in  I.  254  the  point  O  is  a  point  on  the  locus.  Another 
point  of  the  locus  is  obtained  by  finding  Q  such  that  the  per- 
pendiculars QR,  QS  are  respectively  equal  to  A,  B  (I.  257, 
ex.  2). 

Now  the  line  QP  must  pass  through  0,  because  the 
broken  lines  RQS  and  MPN  are  similar  and  similarly  placed 
(72,  ex.  1).  Hence  any  point  P,  situated  in  the  angle  ROS 
and  satisfying  the  given  condition,  lies  on  the  fixed  line  OQ. 

Conversely,  any  point  on  the  line  OQ  satisfies  the  given 
condition.     [The  proof  is  left  to  the  student.] 

Show  that  there  is  another  part  of  the  locus. 

Ex.  Find  a  point  from  which  the  perpendiculars  to  three  given 
lines  shall  have  given  ratios  L-.M:  N.     How  many  solutions  are  there  ? 

139.  Problem  18.  To  find  the  locus  of  a  point  such 
that  its  joins  to  two  given  points  shall  have  a  given 
ratio. 


LOCUS  PROBLEMS 


323 


Let  A  and  B  be  the  given  points,  and  H :  K  the  given  ratio. 
Let  P  be  a  point  such 
that  PA  :  PB  =  H :  K. 

To  find  the  locus  of  P. 

The  points  M  and  iV 
which  divide  AB  inter- 
nally and  externally  in 
the  ratio  H:  K  are  evi- 
dently points  on  the 
locus. 

The  line  PM  bisects 
the  angle  APB,  because  PA:  PB  =  AM :  MB.  [54 

Similarly  PiV bisects  the  external  angle  between  PA  and  PB. 

Then  the  angle  MPN  is  a  right  angle  (I.  100,  ex.  1), 

Hence  P  is  on  the  circle  whose  diameter  is  M]^  (III.  57). 

It  follows  that  any  point  satisfying  the  given  condition 
lies  on  this  circle. 

Conversely,  to  prove  that  every  point  on  this  circle  satis- 
fies the  given  condition. 

Let  P  be  any  point  on  the  circle. 

To  prove  that  PA:PB  =  H:  K. 

Draw  PA'  making  the  angle  MP  A'  equal  to  MPB,  and  meet- 
ing AB  in  some  point  A',  which  is  to  be  proved  coincident 
with  A. 

Since  MPN  is  a  right  angle,  hence  PN  bisects  the  external 
angle  between  PA'  and  PB.  Therefore  A'  is  the  harmonic 
conjugate  of  B  with  regard  to  M  and  N  (59). 

But  A  is  also  the  harmonic  conjugate  of  B  with  regard  to 
M  and  N'.     Therefore  A'  coincides  with  A  (61). 

Hence  PM  bisects  the  angle  APB  ;  and  therefore 
PA:  PB  =  AM :  MB  =H:K. 

Ex.  1,     Compare   the  position   of    the  locus  in   the  three  cases 
H<  =  >K. 

Ex.  2.  Find  the  position  of  a  point,  whose  joins  to  three  given 
points  have  given  ratios  H:  K:L;  show  that  there  may  be  two  soki- 
tions,  one  solution,  or  none. 


324  PLANE  GEOMETRY — BOOK  V 


EXERCISES 

1.  The  rectangle  of  two  lines  is  a  mean  proportional  between  the 
squares  on  the  lines. 

2.  Find  a  line  such  that  the  perpendiculars  to  it  from  three  given 
points  may  have  given  ratios  to  each  other. 

3.  A  regular  polygon  inscribed  in  a  circle  is  a  mean  proportional 
between  the  inscribed  and  circumscribed  circles  of  half  the  number  of 
sides. 

4.  Construct  a  triangle,  being  given  its  base,  ratio  of  sides,  and 
either  altitude  or  vertical  angle. 

6.  Find  the  locus  of  a  point  at  which  two  given  circles  shall 
subtend  equal  angles. 

6.  Two  diagonals  of  a  regular  pentagon  divide  each  other  in 
extreme  and  mean  ratio. 

Definition.  Two  points  P  and  P  are  said  to  be  similarly  situated 
(or  to  correspond)  with  regard  to  two  circles,  whose  centers  are  C  and 
C",  when  CP  and  CF*  are  parallel  and  in  the  ratio  of  the  radii.  The 
correspondence  is  called  direct  or  transverse  according  as  OP  and 
O'P  are  at  the  same  or  opposite  sides  of  the  central  line.  The  fol- 
lowing exercises  illustrate  the  theory  of  correspondence. 

7.  The  line  PP*  divides  the  central  line  either  externally  or  inter- 
nally in  the  ratio  of  the  radii,  and  each  of  the  points  of  division  is  a 
self-corresponding  point  (called  a  center  of  similitude). 

8.  Two  polygons  whose  respective  vertices  are  all  in  direct  (or 
transverse)  correspondence  are  similar  and  have  the  point  S  (or  S')  for 
center  of  similitude. 

9.  The  line  joining  a  fixed  point  to  a  variable  point  on  a  fixed 
circle  is  divided  in  a  constant  ratio ;  prove  that  the  locus  of  the  point 
of  division  is  a  circle,  and  that  the  two  circles  have  the  fixed  point  as 
center  of  similitude. 

10.  In  a  given  sector  OAB  inscribe  a  square,  so  that  two  corners 
may  be  on  the  arc  AB.  [Take  any  square  having  a  side  parallel  to 
AB ;  circumscribe  it  by  a  sector  having  its  radii  parallel  to  OA  and 
OB;  then  use  the  principle  of  correspondence.] 

11.  A  common  tangent  passes  through  a  center  of  similitude. 

12.  Describe  a  circle  through  a  given  point  (P)  to  touch  two  given 
lines  (0^,  OB).  [Draw  any  circle  touching  the  two  lines ;  let  it  meet 
OP  in  P ;  then  considering  P  and  P  as  corresponding  points,  find 
the  center  of  the  required  circle.    Two  solutions.] 


BOOK  VL  — MENSURATION 

1.  Mensuration  is  the  science  of  measurement.  The 
operation  of  measuring  a  magnitude  by  means  of  another 
magnitude  of  the  same  kind  will  be  defined  after  certain 

,  preliminary  notions  are  explained.  It  will  be  shown  to  be 
intimately  connected  with  the  theory  of  ratio  set  forth  in 
Book  IV  and  applied  in  Book  V. 

In  Art.  11  of  Book  IV  the  scale  of  relation  of  two  magni- 
tudes of  the  same  kind  was  explained,  and  it  was  shown 
that  any  two  such  magnitudes  have  a  definite  ascending 
order  in  which  their  various  multiples  occur.  The  theorems 
of  Books  IV  and  V  are  based  on  the  mere  fact  that  this 
scale  is  definite,  and  their  proofs  do  not  require  the  actual 
determination  of  any  particular  scale  of  relation. 

The  determination  of  a  scale  (or  of  a  selected  portion  of 
it)  is,  however,  important  for  other  purposes,  and  is  the 
fundamental  problem  in  Mensuration. 

ABBREVIATED  SCALE 

2.  If  two  magnitudes  A  and  B  are  commensurable,  then 
some  of  their  multiples  are  equivalent  and  occupy  the  same 
place  in  the  ascending  scale  of  magnitude.  If  any  two 
multiple-s  mA  and  nB  are  known  to  be  equivalent,  then  the 
ratio  A:  B  is  completely  defined,  for  it  is  equal  to  the  ratio 
of  two  whole  numbers  n :  m  (IV.  43),  and  hence  the  order 
of  any  assigned  multiples  could  be  written  down. 

Any  ratio  that  is  equal  to  the  ratio  of  two  whole  numbers 
is  called  a  rational  ratio;  thus  the  ratio  of  two  commen- 
surable magnitudes  is  a  rational  ratio,  and  the  ratio  of  two 
incommensurable  magnitudes  is  an  irrational  ratio. 

325 


326  PLANK  GEOMETRY  —  BOOK  VI 

If  two  magnitudes  A  and  B  are  incommensurable,  it  will 
be  proved  presently  that  their  ratio  can  be  sufficiently  char- 
acterized by  assigning  the  intervals  in  which  merely  the 
decimal  multiples  of  the  antecedent  (A,  10  a,  100^,  •••)  are 
found  among  all  the  multiples  of  the  consequent  {B,  25, 
3  By  •••).  Such  an  arrangement  is  called  the  abbreviated 
scale  of  A  and  B. 

The  following  is  an  example  of  an  abbreviated  scale : 

2b,  a,  3b,  •"21b,  10a,  22b,  ...  215 5,  100^,  216^, 
...  21595,  1000^,21605,  .... 

This  exhibits  the  position  of  the  decimal  multiples  of  the 
antecedent,  showing  that 

A  lies  between        2  b  and        3  b, 

10  a  lies  between      21 5  and      225, 

100^  lies  between     215  5  and    216  5, 

1000^  lies  between  2159  5  and  2160  5. 

The  abbreviated  scale  of  /*  and  Q  is  said  to  be  similar  to 
that  of  A  and  5  if  all  the  like  decimal  multiples  of  P  and  A 
lie  between  like  multiples  of  Q  and  5. 

E.g.,  if  the  abbreviated  scale  of  P  and  Q  is 

2q,  P,  3Q,  ...,  21  Q,  10  P,  22  Q,  ..., 

then  the  abbreviated  scale  of  P  and  Q  is  said  to  be  similar 
to  that  of  A  and  5  written  above. 

Similar  scales, 

3.  Theorem  1.  //  the  abbreviated  scale  of  A  and  B 
is  similar  to  tJie  dbbreviojted  scale  of  P  and  Q,  tlien 
the  complete  scales  are  similar,  that  is  to  say,  the 
ratios  A  '•  B  and  P  '■  Q  are  equal. 

For   suppose,  if  possible,  that  the   complete   scales  are 

somewhere  unlike,  then,  by  definition,  the  two  ratios   are 

unequal,  say 

^       '      -^  A'.B^P.Q 


ABBREVIATED   SCALE  327 

Convert  these  ratios  so  as  to  have  a  common  consequent 
T  (V.  17,  114),  and  let  them  become  R :  T  and  S :  r, 
then  R:  T>  S:  T, 

hence  R>S.  [IV.  29 

Divide  T  successively  into  10,  100,  1000,  .••  parts,  until 
a  part  (say  the  one  thousandth)  is  found  which  is  less  than 
the  difference  between  R  and  S.  Take  a  sufficient  number 
(say  m)  of  these  parts,  so  that  m  of  the  parts  shall  be  less 
than  R,  and  not  less  than  S.  Then  R  contains  more  than  m 
thousandths  of  r,  while  S  contains  not  more  than  m  thou- 
sandths of  T.     Hence 

R:T>m:  1000, 

and  S:T-:^m:  1000 ; 

that  is  A:B  >  m  :  1000, 

and  P\Q^m:  1000 ; 

therefore  1000  A  >  mB,                                [IV.  41 

and  1000  P>  7/1 Q.                                 [IV.  42 

Hence  the  thousandth  multiples  of  the  antecedents  occupy 
different  positions  in  the  two  scales. 

Therefore  the  abbreviated  scales  are  unlike,  which  is  con- 
trary to  the  hypothesis.  Hence  the  supposition  made  is 
false;  that  is  to  say,  the  complete  scales  are  everywhere 
alike,  and  A:  B  =  P  :  Q, 

Disshnilar  scales, 

4.  Cor.     If  A:B>P'.Q, 

then  some  decimal  multiple  of  A  occupies  a  more  advanced 
position  among  the  multiples  of  B  than  the  like  decimal  multiple 
of  P  occupies  among  the  multiples  of  Q. 

(This  is  proved  in  the  course  of  the  proof  of  theorem  1.) 

5.  Note.  It  follows  from  3  and  4  that  the  abbreviated 
scale  will  serve  the  same  purpose  as  the  complete  scale,  and 
is  sufficient  to  characterize  the  corresponding  ratio. 

MCM.  ELEM.  GEOM.  — 22 


328  PLANE  GEOMETRY  —  BOOK  VI 

Associated  Numerical  Ratios 

6.  The  abbreviated  scale  may  be  used  to  write  down  two 
sets  of  numerical  ratios,  such  that  the  ratios  of  one  set  are 
each  less  than  the  given  ratio,  and  those  of  the  other  set 
each  greater  than  the  given  ratio. 

E.g.j  from  the  abbreviated  scale  in  Art.  2 
A>2b, 
hence  A:B>2:1,  [IV.  43 

again  A<S  By 

hence  A:B  <S:1. 

In  the  same  way       ^  :  5  >  21 :  10, 
and  A:B<22:10. 

Thus  the  ratio  A:  B  is  greater  than  each  of  the  numerical 
ratios  2 : 1,  21 :  10,  215 :  100,  2159 :  1000,  •.., 

and  less  than  each  of  the  ratios 

3 : 1,  22 :  10,  216  :  100,  2160  :  1000,  .... 

Each  ratio  of  the  first  set  is  called  an  inferior  decimal 
proximate  of  the  given  ratio,  and  each  ratio  of  the  second 
set  a  superior  decimal  proximate.  The  successive  proxi- 
mates  are  said  to  be  of  the  first  order,  the  second  order,  and 
so  on. 

E.g.j  the  ratio  216 :  100  is  the  third  superior  proximate  of* 
the  ratio  A :  B  above. 

A  general  definition  will  now  be  given. 

7.  Definition.  If  a  certain  ratio  lies  between  two  numeri- 
cal ratios  whose  consequents  are  each  equal  to  the  n***  power 
of  10,  and  whose  antecedents  differ  by  unity,  then  the  less 
of  the  two  ratios  is  called  the  inferior  (and  the  greater  the 
superior)  decimal  proximate^  of  the  (n  -\- 1)"'  order,  to 
the  ratio  that  lies  between  them. 

From  this  definition  and  Arts.  3,  4  the  following  corolla- 
ries are  immediate  inferences. 


ABBREVIATED   SCALE  329 

8.  Cor.  I.  If  two  ratios  are  equal,  then  their  corresponding 
decimal  proximates  are  equal. 

9.  Cor.  2.  If  one  ratio  is  greater  than  another,  then  some 
inferior  decimal  j^roximate  of  the  first  is  greater  than  any  infe- 
rior decimal  proximate  of  the  second. 

10.  While  the  use  of  decimal  proximates  is  especially 
applicable  to  irrational  ratios,  it  is  to  be  observed  that 
rational  ratios  also  have  their  inferior  and  superior  decimal 
proximates. 

E.g.,  the  ratio  1 : 3  has  the  inferior  proximates 
3:10,33:100,333:1000,  ..-, 
and  the  superior  proximates 

4:10,34:100,334:1000,  .... 

The  series  of  proximates  to  a  certain  ratio  A :  B  will  termi- 
nate if  it  happens  that  some  decimal  multiple  of  A  is  exactly- 
equivalent  to  some  multiple  of  B. 

E.g.,  if  the  magnitudes  A  and  B  mentioned  above  are  such 
tliat  10000  A  =  21593  B, 

then  A:B  =  21593 :  10000,  [IV.  43 

which  is  both  a  rational  ratio  and  a  decimal  ratio.  This 
ratio  would  be  the  last  of  the  series  of  decimal  proximates 
to  the  ratio  A:  B. 

11.  Definition.  The  ratio  of  any  two  magnitudes  of  the 
same  kind  is  called  a  decimal  ratio  if  it  can  be  exactly 
expressed  as  a  numerical  ratio  whose  consequent  is  a  power 
of  10.  If  it  cannot  be  so  expressed  it  is  called  a  non-deci- 
mal ratio. 

A  non-decimal  ratio  may  be  either  rational  or  irrational. 

12.  Ex.  1.  If  a  given  non-decimal  ratio  is  greater  than  any  other 
given  ratio,  then  some  inferior  decimal  proximate  of  the  first  ratio  is 
greater  than  the  second  ratio.     (The  line  of  proof  is  as  in  Arts.  2,  3.) 

13.  Ex.  2.  If  a  given  non-decimal  ratio  is  less  than  any  other 
given  ratio,  then  some  superior  decimal  proximate  of  the  first  ratio  is 
greater  than  the  second. 


330  PLANE  GEOMETRY  —  BOOK  VI 


NUMBER-CORRESPONDENT 

14.  Definition.  If  the  antecedent  of  a  numerical  ratio  is 
divided  by  its  consequent,  the  quotient  is  called  the  num- 
ber-correspondent of  the  given  ratio,  or  of  any  ratio  equal 
to  it. 

E.g. J  the  ratio  10 :  5  has  the  number-correspondent  2 ;  the 
ratio  5 :  10  has  the  number-correspondent  y^^  or  ^ ;  the  ratio 
9  : 5  has  the  number-correspondent  f . 

If  two  commensurable  magnitudes  A  and  B  have  the 
common  measure  P,  and  if  P  is  contained  m  times  in  A,  and 
n  times  in  5,  then 

A'.B=zmP:nP  =  m:  n.  [IV.  38 

Hence  the  number-correspondent  of  the  ratio  A:  Bis  — • 

n 

Any  number  that  can  be  expressed  as  the  quotient  of  two 
■whole  numbers  is  called  a  rationed^  number. 

Hence  the  number-corresjjondent  of  the  ratio  of  any  two 
commensurable  magnitudes  is  a  rational  number. 

For  this  reason  such  a  ratio  is  called  a  rational  ratio  (2). 

Cofnparlson  of  two  ratios, 

15.  Theorem  2.  According  as  one  rational  ratio  is 
greater  than,  equal  to,  or  less  than  another  rational 
ratio,  so  is  the  number-coi^espondent  of  the  first 
greaiycr  than,  equal  to,  or  less  than  tlie  number- 
correspondent  of  the  second. 

Let  the  two  ratios  be  respectively  equal  to  the  numerical 
ratios  m  :  n  and  p  :  q. 

If  m:n>p:qj 

then  mq  >  np,  [IV.  41 

hence,  by  division,  —  >  — , 

'    "^  '  nq      nq 


NUMBER-C0EBE8P0NBENT  331 

therefore,  by  reducing  the  fractions  to  lowest  terms, 

n      q 

If  the  sign  >  is  replaced  by  either  <  or  ==,  the  proof  is 
similar. 

Addition  of  ratios. 

16.  Theorem  3.  The  number-correspondent  of  the 
sum  of  two  rational  ratios  is  equal  to  the  sum  of 
their  number-correspondents. 

For  the  numerical  ratios  m  :  n  and  p  :  q  are  respectively 
equal  to  the  ratios  mq  :  7iq,  np  :  nq ;  hence  their  sum  is  equal 
to  the  ratio  ^^  ^  ,,^  .  ,,^^  [-y,  115 

whose  number-correspondent  is  — —^  which  equals   the 

sum  of  the  numbers  —  and  -• 
n  q 

17.  Cor.  The  number-correspondent  of  the  difference  of  two 
ratios  equals  the  difference  of  their  number-correspondents. 

Compounding  ratios,      ' 

18.  Theorem  4.  The  ratio  compounded  of  two 
rational  ratios  has  a  number-correspondent  equal  to 
the  product  of  their  nuinber-correspondents. 

For  the  ratio  compounded  of  the  numerical  ratios  m  :  n 
and  p :  q  equals  mp :  nq  (V.  26)  ;   and   the    number-corre- 

spondent  of  this  ratio  is  ^,  which  equals  the  product  of  the 
numbers  —  and  -• 


nq 

and  -• 
n  Q 


19.  It  follows  from  15,  16,  18  that  any  two  rational 
ratios  can  be  compared,  added,  compounded,  etc.,  by  means 
of  their  number-correspondents.  Hence  the  number-corre- 
spondent of  a  rational  ratio  is  sufficient  to  characterize  it. 


332  PLANE  GEOMETRY  —  BOOK   VI 


IRRATIONAL  NUMBERS 

20.  If  A  and  B  are  incommeDsurable,  the  ratio  A :  B  has 
no  rational  number-correspondent.  Such  a  ratio  has  been 
shown,  however,  to  have  two  series  of  proximate  numerical 
ratios,  and  each  proximate  has  its  own  number-correspondent. 
These  number-correspondents  collectively  characterize  the 
ratio. 

By  general  agreement  it  is  usual  to  say  that  the  irrational 
ratio  A :  B  has  then  an  irrational  number-correspondent, 
characterized  or  defined  by  the  two  categories  of  rational 
numbers,  the  decimal  proximates,  just  as  the  ratio  itself  is 
characterized  or  defined  by  the  order  of  certain  multiples. 

These  categories  of  rational  numbers  are  called  the  two 
decimal  categories  belonging  to  the  irrational  niunber. 

The  number-correspondent  of  any  ratio  A:  B  is,  denoted  by 

the  symbol  — 

It  is  now  necessary  to  give  definitions  of  the  words  equal, 
greater,  less,  sum,  product,  etc.,  when  applied  to  the  irra- 
tional numbers  just  defined.  The  definitions,  and  cTertain 
inferences  from  them,  are  given  in  the  following  articles. 

21.  Definition.  An  irrational  number  is  said  to  be  greater 
than,  equal  to,  or  less  than  another  number  (whether  rational 
or  irrational)  according  as  the  ratio  to  which  the  first  number 
corresponds  is  greater  than,  equal  to,  or  less  than  the  ratio 
to  which  the  second  number  corresponds. 

22.  Theorem  2  may  now  be  restated  without  restriction  : 

According  as  one  ratio  is  greater  than,  equal  to, 
or  less  than  another  ratio,  so  is  the  number-corre- 
spondent of  the  first  greater  than,  equal  to,  or  less 
than  the  numher-correspondent  of  the  second. 

A  C 

I.e.,  according  as  A  -.  B  >  =  <  c :  Dy  so  is  —>  =  <-. 

B  D 


IBRATIONAL  NUMBERS  333 

23.  Cor.  I.  When  two  irrational  numbers  are  equal,  their 
decimal  categones  are  identical,  respectively. 

For  the  abbreviated  scales  of  their  corresponding  ratios  are  similar 
(def.  and  3),  hence  the  decimal  proximates  are  alike  (6). 

24.  Cor.  2.  If  two  irrational  numbers  have  identical  deci- 
mal categories,  then  the  irrational  numbers  are  equal. 

For  then  the  decimal  proximates  are  alike,  hence  the  abbreviated 
scales  are  alike  (6),  and  hence  the  corresponding  ratios  are  equal. 

25.  Cor.  3.  If  one  irrational  number  is  greater  than  an- 
other, then  some  inferior  decimal  proximate  of  the  first  is 
greater  than  any  inferior  decimal  proximate  of  the  second  (9). 

Ex.  1.  If  an  irrational  (or  a  non-decimal  number)  is  greater  than 
any  other  given  number,  then  some  inferior  decimal  proximate  of  the 
first  number  is  greater  than  the  second  (12). 

Ex.  2.  If  an  irrational  number  (or  a  non-decimal  number)  is  less 
than  any  other  given  number,  then  some  superior  decimal  proximate 
of  the  first  number  is  less  than  the  second  (13). 

26.  Definition.  The  suTn  of  two  irrational  numbers  is 
defined  as  the  number-correspondent  of  the  ratio  which  is 
the  sum  of  the  two  ratios  corresponding  to  the  two  irra- 
tional numbers. 

A  similar  definition  applies  to  the  difference  of  two  irra- 
tional numbers,  and  also  to  the  sum  (or  difference)  of  a 
rational  and  an  irrational  number. 

27.  Theorem  3  may  now  be  restated  without  restriction  : 

The  numher-correspofident  of  the  sum  of  any  two 
ratios  is  equal  to  the  sum  of  their  number-corre- 
spondents. 

28.  Cor.  I.  The  sum  of  two  irrational  numbers  is  greater 
than  the  sum  of  any  two  numbers  that  are  inferior  proximates 
to  them,  respectively,  and  less  than  the  sum  of  any  two  supenor 
^proximates.     (Use  22,  27  ;  and  V.  120.) 


334  PLANE  GEOMETRY  —  BOOK  VI 

Addition  cormnutaMve. 

29.  Cor.  2.  Tlie  addition  of  numbers  is  a  commutative 
operation,  i.e.  the  sum  of  any  two  or  more  numbers  is  the 
same,  in  whatever  order  they  may  be  taken  (V.  118). 

30.  Definition.  The  product  of  two  irrational  numbers 
(or  of  a  rational  number  and  an  irrational  number)  is  defined 
as  the  number-correspondent  of  that  ratio  which  is  com- 
pounded of  the  ratios  corresponding  to  the  given  numbers. 

31.  Theorem  4  may  now  be  stated  without  restriction : 

The  ratio  compounded'  of  any  two  or  more  ratios 
ha^  a  number-correspondent  equal  to  the  product  of 
their  nuviber-correspondents. 

The  process  of  finding  the  product  of  two  or  more  num- 
bers is  called  multiplication. 

Multiplication  commutative. 

32.  Cor.  I.  Multiplication  is  a  commutative  operation,  i.e. 
the  product  of  any  two  or  more  numbers  is  the  same,  in  what- 
ever order  they  may  be  taken.     (Use  definition,  and  V.  27.) 

Multiplication  distributive. 

33.  Cor.  2.  Multiplication  is  distributive  as  to  addition, 
i.e.  the  product  of  any  number  by  the  sum  of  any  other  num- 
bers is  equal  to  tlie  sum  of  the  products  of  the  first  number 
by  the  other  numbers  separ^ely  (V.  121). 

MEASURE-NUMBER 

34.  Definitions.  The  ratio  which  any  magnitude  bears 
to  a  standard  magnitude  of  the  same  kind  is  called  the 
measure-ratio  of  the  first  magnitude. 

The  number-correspondent  of  the  measure-ratio  of  any 
magnitude  is  called  the  measure-number  of  that  magni- 
tude. 


MEASURE-NUMBER  335 

The  measure-number  of  a  magnitude  is  rational  or  irrar 
tional  according  as  the  magnitude  is  or  is  not  commensur- 
able with  the  standard  magnitude  (14). 

If  M  is  any  magnitude,  and  S  the  standard  magnitude  of 
the  same  kind,  then  the  measure-ratio  of  M,  and  the  measure- 
number  of  M,  are  respectively 

M 

if:  Sand-. 
S 

The  measure-number  of  a  straight  line  is  called  its  length 
with  reference  to  the  standard  line. 

The  measure-number  of  a  polygon  is  called  its  area  with 
reference  to  the  standard  polygon. 

The  square  described  on  the  standard  line  is  usually  taken 
as  the  standard  polygon. 

The  universal  standard  of  line-magnitude  adopted  by 
scientific  men  is  the  meter.  It  is  the  largest  dimension 
of  a  certain  standard  bar  of  platinum  when  taken  at  the 
temperature  of  melting  ice.  This  standard  bar  is  carefully 
preserved  in  the  Paris  observatory. 

The  first  three  decimal  multiples  of  the  meter  are  denoted 
by  prefixes  formed  from  the  Greek  words  for  10,  100,  1000. 


Name 

Magnitude 

Abbreviation 

meter 

standard 

m. 

dekameter 

10  meters 

Dm. 

hektometer 

100  meters 

Hm. 

kilometer 

1000  meters 

Km. 

The  first  three  decimal  submultiples  of  the  meter  are 
denoted  by  prefixes  formed  from  the  Latin  words  for  10, 
100,  1000. 

Name  Magnitude  Abbreviation 

decimeter  one  tenth  meter  dm. 

centimeter         one  hundredth  meter  cm. 

millimeter         one  thousandth  meter  mm. 


336 


PLANE  GEOMETRY  —  BOOK  VI 


=. —  OD 


A  straightedge  on  which  are  marked  divisions  equal  to  a 
meter  and  to  its  decimal  submultiples  is  called  a  measur- 
ing'line. 

With  such  a  measuring-line  the  succes- 
sive decimal  proximates  to  the  measure- 
number  of  any  other  accessible  line  can 
be  found  as  follows : 

Apply  the  meter  in  succession  as  often 
as  it  will  go  until  the  remainder  is  less 
than  a  meter.  Suppose  the  meter  goes 
3  times.  Then  3  is  the  first  inferior 
proximate,  and  4  the  first  superior  proxi- 
mate, to  the  measure-number  of  the  given 
line. 

Next,  to  the  remainder  apply  the  deci- 
meter as  often  as  it  will  go  until  there 
is  a  remainder  less  than  a  decimeter. 
Suppose  the  decimeter  goes  5  times. 
Then  the  proximates  of  the  second  order 

3  +  ^,3  +  A. 


I 


^^00 


Again,  to  the  last  remainder  apply  the 
centimeter  until  the  remainder  is  less 
than  a  centimeter.  Suppose  it  goes  8 
times.  Then  the  proximates  of  the  third 
order  are 

3  +  A  +  T*tr.  S  +  A  +  T^TT- 

Next,  to  the  last  remainder  apply  the 

millimeter,  and    suppose    it    goes    twice 

with  a  remainder  less  than  a  millimeter. 

Then  the  proximates  of  the  fourth  order 

are 

S  +  A  +  yfr  +  TAir,  3-f^-f-rf 


+  1000* 


;o 


-LD 


to 


CVJ 


Again,  to  the  last  remainder  apply  the  tenth  of  the  milli- 
meter;   suppose  it  goes  four  times  with  a  remainder  less 


MEASUREMENT  OF  RECTANGLES 


337 


than  the  divisor.  Then  the  proximates  of  the  fifth  order 
are 

^  +  A  +  TTo  +  T oV^  +  loooo  ^  +  A  +  yfo  +  Tihu  +  Tnhfdy 

or,  in  the  decimal  notation,  3.5821,  3.5822.  The  error  of 
either  of  these  last  proximates  is  less  than  one  tenth  of  a 
millimeter,  i.e.  one  ten-thousandth  of  a  meter. 


MEASUREMENT  OF  RECTANGLES 

35.  Theorem  5.  If  the  standard  polygon  is  the 
square  described  on  the  standard  line,  then  the 
measure-numher  of  a  rectangle  equals  the  product 
of  the  measure-numhers  of  two  adjacent  sides. 

Let  I  be  the  standard  line ;  S  the  standard  square  whose 
side  \^l\  R  the  rectangle  whose  adjacent  sides  are  the  lines 
a  and  h. 

The  surface-ratio  R :  S  equals  the  ratio  compounded  of  the 
line-ratios  a :  I  and  b:l  (V.  100). 

Therefore,  by  31,  the  number-correspondent  oi  R:  S  equals 
the  product  of  the  number-correspondents  of  a :  ^  and  b  il-^ 
that  is  to  say  „  , 

i.e.  the  measure-number  of   R  equals  the   product  of  the 
measure-numbers  of  its  adjacent  sides  a  and  b. 

Ex.  1.  Find  the  measure-number  of  a  rectangle  whose  sides  are  3 
and  4  centimeters  respectively. 

Taking  the  centimeter  as  stand- 
ard line  and  the  square  centimeter 
as  standard  surface,  it  is  evident 
from  the  figure  that  the  measure- 
number  of  the  rectangle  is  12,  which 
agrees  with  the  theorem.  This 
method  of  proof  does  not  apply 
when  either  of  the  sides  is  incom- 
mensurable with  the  standard  line. 


338  PLANE  GEOMETRY  —  BOOK  VI 

Ex.  2.  Find  the  measure-number  of  a  rectangle  whose  sides  are 
2  meters  and  1  decimeter. 

In  terms  of  the  meter  the  sides  are  2,  ^^ ;  hence  the  area  equals 
2-rff>  or  ^  of  the  square  meter. 

When  the  decimeter  is  used  as  standard  line,  the  measure-numbers  of 
the  sides  are  20,  1 ;  and  the  area  equals  20. 1,  or  2*0  square  decimeters. 

Ex.  3.  The  sides  of  a  rectangle  are  2.21  m.  14  cm. ;  find  its  area. 
Answer,  .3094  sq.  m.,  or  30.94  sq.  dm.,  or  3004  sq.  cm. 

Ex.  4.  A  rectangle  contains  3  sq.  ra.,  one  side  is  6  cm.,  find  the 
other  side. 

Ex.  6.  What  theorem  in  Book  II  corresponds  to  the  following 
algebraic  theorem  :  a  (b  +  c  +  d)  =  ah  -^  ac  -^  ad? 

36.  Cor.  I.  The  mecisure-number  of  a  square  equals  the 
second  power  of  the  measure-number  of  its  side. 

37.  Note.  For  this  reason  the  second  power  of  a  num- 
ber is  often  called  its  square  ;  and  the  number  whose  second 
power  is  equal  to  the  given  number  is  called  the  square  root 
of  the  given  number. 

Ex.  State  what  theorems  in  Book  II  correspond  to  the  following 
algebraic  theorems :  a{a -\- h)  =  a^  -^  ah\  (a  +  6)'^  =  a2  _|_  52  ^  2  ah. 

38.  Cor.  2.  Tlie  measure-nii^mher  of  the  side  of  a  square 
equals  the  square  root  of  the  measure-number  of  the  square 
itself 

Ex.  A  square  contains  two  square  meters,  find  its  side.  Answer, 
V2  =  1.4142...  m. 

39.  Cor.  3.  In  a  right  triangle  the  measure-number  of  the 
hyi)otenuse  equals  the  square  root  of  the  sum  of  the  squares 
of  the  measure-numbers  of  the  other  two  sides;  and  the 
measure-number  of  one  of  the  perpendicular  sides  equals  the 
square  root  of  the  difference  of  the  squares  of  the  mea.sure- 
numbers  of  the  other  two  sides  (II.  61). 

In  symbols,  if  the  lengths  of  the  perpendicular  sides  are  a,  b,  and 
of  the  hypotenuse  c,  then  c^  =  a^  -\-  b"^,  a^=c^  —  h^. 


DIRECTED  LINES  339 


DIRECTED   LINES 


40.  The  line  joining  two  points  A  and  B  may  be  regarded 
as  reaching  either  from  ^  to  5  or  from  i?  to  ^.  A  segment 
having  the  initial  point  A  and  the  terminal  point  B  is  denoted 
by  AB,  and  the  segment  having  the  initial  point  B  and  the 
terminal  point  A  is  denoted  by  BA. 

The  two  segments  AB  and  BA  are  said  to  be  equal  in 
magnitude  and  opposite  in  direction  or  sense. 

Any  two  collinear  segments  AB  and  CD  may  be  compared 
by  imagining  CD  to  slide,  without  turning  out  of  its  line, 
until  the  initial  point  C  falls  on  the  initial  point  A.  If  the 
terminal  points  are  then  on  the  same  side  of  the  common 
initial  point,  the  two  segments  are  said  to  have  the  same 
sense.     If  not  they  are  said  to  have  opposite  sense. 

Similarly  any  indefinite  line  may  be  regarded  as  traced  in 
either  of  two  opposite  senses  or  directions.  The  sense  in 
which  it  is  supposed  to  be  traced  is  indicated  by  the  order 
of  naming  its  leading  letters. 

Any  segment  of  a  directed  indefinite  line  is  called  a 
forward  or  a  hachward  segment  according  as  its  sense  is 
similar  or  opposite  to  that  of  the  indefinite  line. 


U  A  B  L 

For  instance,  AB  is  a  forward  segment  of  the  line  L'L,  and 
BA  is  a  backward  segment. 

All  forward  segments  of  the  same  or  different  lines  are 
said  to  be  of  the  same  quality,  and  so  are  all  backward 
segments ;  but  any  forward  segment  and  any  backward  seg- 
ment are  said  to  be  of  opposite  quality. 

The  ratio  of  any  two  segments  of  opposite  quality  will  be 
represented  by  a  negative  number. 

A  forward  segment  is  commonly  taken  as  the  standard, 
and  then  any  forward  segment  has  a  positive  measure- 
number,  and  any  backward  segment  has  a  negative  measure- 
number. 


340  PLANE  GEOMETRY  —  BOOK  VI 

The  distance  from  a  point  A  to  another  point  B  is  defined  as  the 
measure-number  of  the  segment  AB. 

A  point  on  a  directed  indefinite  line  is  said  to  divide  it  into  a  for- 
ward part  and  a  backward  part,  which  are  distinguished  by  the  fact 
that  a  segment  reaching  from  any  point  of  the  latter  to  any  point  of 
the  former  is  a  forward  segment.  If  two  parallel  directed  lines  are 
cut  by  a  transversal,  and  if  their  forward  parts  are  at  the  same  side 
of  the  transversal,  then  the  parallels  are  said  to  be  similar  in  direc- 
tion ;  but  if  the  forward  parts  are  at  opposite  sides  of  the  transversal, 
then  the  parallels  are  said  to  be  opposite  in  direction. 

Addition  of  segments* 

41.  Two  collinear  segments  are  added  by  sliding  one  of 
them  so  that  its  initial  point  falls  on  the  terminal  point  of 
the  first.  The  segment  reaching  from  the  initial  point  of  the 
first  to  the  terminal  point  of  the  second  is  called  the  sum 
of  the  two  segments. 

From  this  definition  it  follows  that  the  sum  of  the  col- 
linear segments  AB  and  BC  is  AC^  no  matter  in  what  order 
the  three  points  come  on  the  line. 


B 


The  measure-number  of  the  segment  AB  will  be  denoted 
by  the  symbol  AB. 

Hence,  AB  and  BA  have  opposite  algebraic  signs. 

That  is,  AB  =—BA-^   BA=  —AB. 

Addition  of  tneasure-ninnhers, 

42.  The  meaning  just  given  to  the  addition  of  segments 
corresponds  to  the  algebraic  addition  of  their  measure- 
numbers. 

E.g.,  if     AB  =  7  and  BC  =  "3,  then  AC  =  1  ■\-  ("3)  =  4. 

This  principle  may  be  stated  in  general  terms  thus : 


MEASUREMENT  OF  TRIANGLES  341 

The  sum  of  two  or  more  collinear  segments  has  a 
measure-number  equal  to  the  algebraic  sujn  of  the 
measure-num,bers  of  the  several  segments. 

E.g.y  AB  +  BC  +  CD  =  AD, 

AB  +  BC-^  CA  =  AA^  0. 
A  segment  is  subtracted  by  adding  its  opposite. 
E.g.j        AC  —  BG  =  AC -\-  CB  =  AB, 

DA  —  DB  =  DA  +  BD  =  BD  +  DA  =  BA. 

MEASUREMENT  OF   TRIANGLES 

43.  Algebraic  relations.  One  advantage  of  the  conven- 
tions just  laid  down  is  that  by  taking  account  of  the  sense 
of  collinear  segments,  two  different  geometric  theorems  can 
often  be  made  to  correspond  to  one  algebraic  statement. 
This  is  illustrated  in  some  of  the  following  examples : 

Ex.  1.  The  lengths  of  the  sides  of  a  triangle  are  8,  10,  5  ;  find  the 
segments  of  the  base  made  by  the  perpendicular  to  the  third  side  from 
the*opposite  vertex,  and  also  the  length  of  this  perpendicular. 

When  the  angle  ACB  is  obtuse,  the  relation  between  the  measure- 
numbers  of  the  sides  and  of  the  projections  is  furnished  by  II.  62,  of 
which  the  corresponding  algebraic  statement  is 

AB^  =  AG^  +  BG^  +  2  AG'  GD, 

in  which  AB^  stands  for  the  second  power  of  the  measure-number  of 
the  side  AB,  and  AG  -  GD  for  the  product  of  the  measure-numbers 
of  the  lines  AG  and  GD,  this  product  being  the  measure-number  of 
the  rectangle  contained  by  these  two  lines. 

Again,  when  the  angle  AGB  is  acute,  the  appropriate  relation  is 
furnished  by  II.  63,  of  which  the  corresponding  algebraic  statement  is 

AB^  =  AG-^  +  BG-^  -2  AG-  DG. 

Now  it  will  be  seen  that,  when  account  is  taken  of  the  sense  of  the 
segments  GD  and  DG,  the  two  algebraic  statements  are  identical,  for 
the  second  could  be  derived  from  the  first  by  replacing  GD  by  its 
equivalent  —  DG.  Either  of  these  equivalent  statements  may  be  taken 
to  apply  to  all  cases,  attention  being  paid  to  the  proper  signs  to  be 
given  to  the  segments  GD,  DG,  and  AG.    We  shall  use  the  latter  form, 


342  PLANE  GEOMETRY  —  BOOK  VI 

and  shall  take  ^C  as  positive.  Then  DC  ia  positive  when  D  and  A 
are  at  the  same  side  of  C ;  and  DC  is  negative  when  Z>and  A  are  at 
opposite  sides  of  C. 

When  AB,  AC,  and  5 Care  given,  then  i>C  is  found  correctly  both 
in  magnitude  and  sign  by  solving  the  above  equation. 

Then  AD,  the  other  segment  of  the  base,  is  found  from  the  equa- 

^>^"  AD -\- DC  =  AC, 

which  is  true  irrespective  of  the  order  of  the  points  A,  D,  C. 

Substituting  the  numbers  given  above,  DO  is  to  be  found  from  the 
equation  52  =  8^  +  lO*  -  2  •  10  •  DC. 

Hence  DC  =  '^  +  ^^  "  ^^  =  6.95  m. 

20 

It  may  be  observed  that  since  DC  and  AC  have  the  same  sign, 
hence  D  and  A  are  at  the  same  side  of  C,  and  the  angle  ACB  is  acute. 

Since  AD  +  6.95  =  10,  hence  AD  =  3.06  m. 

Again,  the  altitude  BD  is  given  by 


BD  =  VBC^  -  DC^  =  \/64  -  48.3025 


=  Vl5.6975  =  3.962  m. 
And  the  area  is  given  by  A  =  ^  ^C  •  BD  =  19.81  sq.  m. 

Ex.  2.     The  lengths  of  the  sides  of  a  triangle  are  a,  b,  c ;  find  the 
perpendicular  to  the  side  h  and  the  area.    As  in  the  last  example, 

c2  =  a2  +  ^'2  _  2  &  .  DC, 

hence  DC  =  «^  +  -^i^^ 
26 

and     DD2  =  a2-(«^  +  ^^-^^)^  =  ^«^^^-(«^+^^-^^)' 
4  62  4  62 

^  (2a6  +  a2  +  62  -  c^)  (2ab-a^-I^  +  c^) 
4  62 

^  [(g  -r6)2  -  c2]  [c2  -(a-  6)2] 
4  62 

_  (a+6+c)  (a  +  6-c)  (c+a-6)  (c-a+b) 
4  62 

Now  let  s  be  the  semi-perimeter  ;  then  a  +  6  -f  c  =  2  s,  and  a  +  6  -  c 
=  2(s  -  c).     Similarly  c  +  a  -  6  =  2(s  -  6),  c  -  a  +  6  =  2(s  -  a). 

Hence  ^^.^  16.(5  -  a)  (s  -  6)  (.  -  c)^ 

4  62 


MEASUREMENT  OF  TRIANGLES  343 

2 

and  BD  =  -  Vs(s  -  a)  (s  -  6)  (s  -  c). 

This  is  the  length  of  the  perpendicular  on  the  side  6.    The  other  two 
perpendiculars  can  be  written  down  by  algebraic  symmetry. 
The  area  is  found  from  the  relation 


A=lb-BD=  Vs(s  -a)  (s  -  b)  (s  -  c). 

Thus  the  area  of  a  triangle  equals  the  square  root  of  the  continued 
product  of  the  semi-perimeter  and  the  differences  between  the  semi- 
perimeter  and  each  side  in  turn.     [Heron's  Rule  (110  b.c.).] 

Ex.  3.  The  lengths  of  the  sides  of  a  triangle  are  a,  &,  c.  Find  the 
lengths  of  the  three  medians. 

In  the  figure  of  II.  67,  let  the  lengths  of  the  sides  opposite  the  angles 
A,  B,  C,  be  a,  &,  c ;  and  let  the  length  of  the  median  BD  be  m. 


then  a2  +  c2  =  2m2  +  2Z)C=2  =  2m2  +  2 


(!)• 


Therefore  m  =  lV2a^ +  2  c^  -  b^. 

By  algebraic  symmetry  the  other  two  medians  are 


^V2  62  +  2  c2  -  a2,        ^V2a2  +  262_c2. 

Ex.  4.     Find  the  lengths  of  the  three  bisectors  of  the  angles. 
The  bisector  CD  of  the  angle  C  divides  c  in  the  ratio  a:b  ;  hence 
the  segments  are 

AD  =  -^^— .  c,        DB 


a-h  b  a  +  b 

Now  AC  '  CB  =  AD  •  DB  +  CD^  ;  [V.  127 

hence  ab  =      ^^ 


(a  +  by 

therefore  02)2  _  ^^fi  _ 

L        (a  +  &)2. 

The  lengths  of  the  two  other  bisectors  can  be  written  by  symmetry. 

Ex.  5.    Find  the  radius  of  the  circumscribed  circle. 

From  V.  125,  a-6  =  2i2.j9, 

thus  j^^^^abc^abc 

2p     2pc     4  A 

where  A  stands  for  the  area  of  the  triangle. 

Ex.  6.     Find  the  radius  of  the  inscribed  circle. 

Let  O  be  the  center  of  the  inscribed  circle  and  r  its  radius. 

.Then  OAB  +  OAC  +  OBC  =  ABC, 

McM.   ELEM.  GEOM. — 23 


344  PLANE  GEOMETRY  — BOOK   VI 

hence  ^r  •  AB  +  ^r  -  AC -\- ^r  -  BC  =    A,  [UI.  101 

riAB  +  AC-{-BC)=2Ay 

2  A 

therefore  r  = 

a  +  b  +  c 

Ex.  7.    Prove  that  the  radii  of  the  escribed  circles  are 

--       2A        ,^^      2A      ^^^^      2A      .       ^i,i_jo3 


b  +  c  —  a  a  —  b  +  c  a-^b  —  c 

Ex.  8.  To  compute  the  two  parts  of  a  Hue  wliose  length  is  a,  when 
divided  in  extreme  and  mean  ratio. 

In  the  figure  of  II.  89,  let  AB  -  a.    Then  BE^iyZ-a; 

and  AP  =  K>/6  -  l)a ;  PB  =  K3  -  V5)a. 

MEASUREBIENT  OF  REGULAR  POLYGONS 

44.  General  relations.  Take  a  circle  of  radius  r ;  and  let 
the  sides  of  the  regular  inscribed  and  circumscribed  ?i-gons 
be  denoted  by  s„,  S^ ;  and  the  apothem  of  the  former  by  a^. 
(No  special  symbol  is  needed  for  the  apothem  of  the  latter, 
since  it  is  always  equal  to  r.) 

Among  the  four  numbers,  s„,  s„,  a^,  r,  there  are  two  simple 
general  relations : 

Since  the  apothems  of  two  regular  w-gons  are  in  the  ratio 
of  similitude,  hence  „      „ 

and,  since  the  apothem  bisects  the  side  perpendicularly, 
hence  r,  a„,  i  s„,  are  the  lengths  of  the  sides  of  a  right  tri- 
angle, therefore  ^^  =  a\-{-^  s^^.  (2) 

46.  Special  relations.  In  the  case  of  the  simpler  polygons, 
the  figure  usually  furnishes  some  special  relation  between 
two  of  the  lengths  s„,  a„,  S„,  r.  This  special  relation  together 
with  the  two  general  relations  stated  above  will  be  sufficient 
to  express  any  three  of  these  lengths  in  terms  of  the  fourth. 

In  the  following  examples,  .s„,  a„,  S„,  are  each  expressed  in 
terms  of  r ;  the  values  of  n  are  taken  in  order  of  simplicity. 


MEASUREMENT  OF  REGULAR   POLYGONS       345 

Ex.  1.  For  w  =  6:  show  from  a  figure  that  SQ  =  r;  hence,  by  (2), 
that  ae  =  i  V3  r ;  and,  by  (3),  that  Se  =  f  V3  r. 

Ex.  2.  For  w  =  4  :  show  from  a  figure  that  a^  =  |  S4  ;  hence,  by 
(2),  that  S4  =  V2  r  ;  also  that  S^  =  2r. 

Ex.  3.  For  n  =  S:  show  that  as  =  | r  ;  S3  =  VS r  ;  iSb  =  2\/3 r. 

Ex.  4.   For  n  =  10  :  show  (III.  122,  V.  98,  and  VI.  43,  ex.  8)  that 

Sio=^{V6-l)  r  ;  hence  that  aio=iVlO+2  V5  •  r,  Sio=  ^{^^  - '^)  .  r. 

V1O+2V5 
Ex.  5.  For  n  =  5  :  show  from  the  figure  of  III.  122  that  BK  bisects 
AP  perpendicularly,  and  2  ^5  =  r  +  sio ;  hence  that  as  =  i  (^5  +  1)  r, 

Vs  +  i 

Ex.  6.  For  w  =  16 :  in  figure  of  III.  128,  ^O  =  Se,  AB  =  sio, 
BC  =  Sis.  Let  AO  meet  circle  again  in  D.  Prove  BD  =  2aio, 
CD  =  2  ae,  ^Z>  =  2  r.  Show  by  V.  128  that  2  r  •  sis  +  2  ae  •  Sio 
=  2  aio  •  Sq,  and  hence  that 

si5  =  (aiose  -  aesio)  --=k  [ VlO  +  2 V5  -  V3  ( V5  -  1)]  •  r. 
r 

Then  show  how  to  find  ais,  Sis. 

46.  Regular  2  n-gon.  The  next  step  is  to  show  how  to 
proceed  from  any  of  the  above  regular  polygons  to  another 
of  double  the  number  of  sides. 

Let  s.2n,  S2n,  be  the  sides  of  the  inscribed  and  circumscribed 
2  w-gon ;  then  by  V.  136,  137,  the  following  two  relations 
exist  between  the  four  numbers  s„,  S,^,  Sg^,  S2n  - 


Sr.  S 


(3) 
(4) 


^2n  "2"  ^2» 

I.e.  ^m'-^2«  '=  -^  ^'2n' 

^2n  ^'' 

eliminating^,,,  r  •  .„  =  2  .,,, .  a,„.  (5) 

This  can  also  be  proved  directly  from  the  figure  of  V.  137. 


Now  from  (1),  Art.  44,     -~  =  ~,  whence  (4)  becomes,  by 


346  PLANE  GEOMETRY — BOOK  VI 

Ex.  7.  For  2n  =  8:  put  n  =  4  in  (3),  and  solve  for  8%.  Use 
the  values  of  s^  and  8^  found  in  ex.  2.  The  reduced  result  is 
6^8  =  2(V2  -  1) -r.     Then  show,  from   (4),    that  sg  =  V2  -  V2  •  r, 

/         /~ 

and,  from  (1),  that  a%  = ^^ •  r. 

2(>/2-l) 


Ex.  8.   Show  that  S12  =  V  2  -  \/3  .  r. 

47.  Apothem  in  terms  of  side.  It  is  often  convenient  to 
know  the  value  of  a^  in  terms  of  s^.  In  the  above  examples 
a„  and  s^  are  each  expressed  in  terms  of  r.  Hence  a„  can 
be  expressed  in  terms  of  s^,  when  n  is  3,  4,  5,  6,  8,  etc. 

a3=iV3.Ss;  a^=^^s^',  a8=iV3.S6;  Og  =  i(V2 +  I)s8; 


_v^+j_  _2VlO  +  2V5 

2V 10  -  2 V5  V5  -  1 

48.  Area  in  terms  of  side.  A  regular  w-gon  is  equivalent 
to  the  sum  of  n  triangles,  each  having  its  base  equal  to  the 
side,  and  its  altitude  equal  to  the  apothem.  Hence  the  area 
.„  is  given  by  ^.  =  j  „.,.„.. 

Therefore,  by  47, 

^«  =  (V2  +  l).s^^=    5(V5  +  1)    ,.,    • 
4V10  -  2V5 

Ex.  The  regular  pentagon  is  about  1.72  times  the  square  on  its  side. 
The  regular  hexagon  is  about  2.6  times  the  square  on  its  side. 


MEASUREMENT   OF   THE   CIRCLE 

49.  Hitherto  we  have  been  concerned  with  the  measure- 
ment of  figures  bounded  by  straight  lines.  To  lead  up  to 
the  measurement  of  the  circle,  it  is  necessary  to  give  some 
elementary  principles  relating  to  variables  and  their  limits. 


MEASUREMENT  OF  THE  CIRCLE  347 

Variables  and  Limits 

50.  Definitions.  A  number  which  takes  a  series  of  differ- 
ent values  in  succession  is  called  a  variable. 

E.g.,  the  population  of  a  city  in  successive  years;  the 
number  of  seconds  between  sunrise  and  sunset  on  succes- 
sive days ;  the  perimeter  of  a  regular  polygon  inscribed  in 
a  given  circle,  when  the  number  of  sides  is  3,  4,  5,  6,  •••  in 
succession. 

When  the  law  of  change  of  a  variable  is  such  that  its 
successive  values  approach  nearer  and  nearer  to  a  certain 
fixed  number  so  that  the  difference  between  the  latter  and 
the  variable  can  become  and  remain  smaller  than  any 
assigned  number,  then  the  fixed  number  is  called  the 
limit  of  the  variable  in  question. 

E.g.,  the  series  of  fractions,  \,  f,  f,  |,  -f^,  W,  •••  (in 
which  each  term  is  derived  from  the  preceding  by  add- 
ing two  units  to  numerator  and  denominator),  approaches 
unity  as  a  limit;  for  by  continuing  the  series  far  enough 
under  the  same  law,  a  term  will  be  reached  that  differs 
from  unity  by  less  than  any  assigned  number,  however 
small ;  for  instance,  if  the  assigned  number  is  y^Vttj  ^^  ^^^ 
continue  the  series  up  to  the  term  \%^,  which  differs  from 
unity  by  less  than  the  assigned  number. 

Again,  the  series  of  numbers  8,  4,  2,  1,  i  \,  \i  ^,  •••  (in 
which  each  term  is  half  the  preceding),  tends  toward  zero 
as  a  limit;  for  the  series  can  be  continued  until  a  term  is 
reached  which  is  less  than  any  assigned  number. 

Two  variables  are  said  to  be  related  when  one  depends 
on  the  other,  so  that  when  the  value  of  one  is  known,  the 
value  of  the  other  can  be  found. 

E.g.,  the  length  of  the  side  of  a  square  and  its  area  are 
related  variables.  If  the  side  takes  the  series  of  values 
1,  2,  3,  4,  •••,  then  the  area  takes  the  series  of  values  1,  4, 
9,  16,  •••,  each  term  in  the  latter  series  being  the  second 
power  of  the  corresponding  term  in  the  former  series. 


348  PLANE  GEOMETRY  —  BOOK   VI 

ELEMENTARY    PRINCIPLES    OF    LIMITS 

51.  Principle  1.  If  two  variable  numbers  are  so 
related  that  they  remain  always  equal,  and  if  one  of 
them  approaclves  a  limit,  then  the  other  approaxihes 
the  same  limit. 

For  the  two  equal  variables  are  at  all  stages  represented 
by  one  number,  and  this  number  has  only  one  limit. 

52.  Principle  2.  If  two  finite  related  variables  are 
such  that  their  quotient  apj)roaches  unity  as  a  limit, 
then  their  difference  approaches  zero  as  a  limit. 


Let  the  two  variables  be  represented  by  the  measure- 
numbers  of  the  finite  lines  OA  and  OB. 

By  hypothesis,  the  measure-number  of  the  ratio  OA  :  OB 
approaches  unity  as  a  limit ;  hence  the  points  A  and  B  can 
come  as  near  together  as  desired.  Therefore,  the  difference 
of  the  two  variables  approaches  zero  as  a  limit. 

53.  Cor.  If  two  finite  related  variables  are  such  that  their 
difference  approaches  zero  as  a  limity  then  their  quotient  ap- 
proaches unity  as  a  limit. 

54.  Principles.  If  a  variable  continually  increases, 
and  never  exceeds  a  certain  fixed  number,  then  the 
variable  approa/ihes  some  limit  not  greater  than  the 
fixed  number. 

For  if  the  variable  has  no  limit,  it  must  (since  it  continu- 
ally increases)  ultimately  exceed  any  assigned  number. 

55.  Cor.  If  a  variable  continually  decreases,  and  never  be- 
comes less  than  a  certain  fixed  number^  then  the  variable 
approaches  some  limit,  not  less  than  the  fixed  number. 

Ex.  If  a  variable  increases  toward  a  certain  limit,  and  if  the 
variable  always  exceeds  a  certain  fixed  number,  then  the  limit  exceeds 
this  fixed  number. 


-t — I 1 


MEASUREMENT  OF  THE  CIRCLE  349 

56.  Principle  4.  If  one  of  two  related  variables  is 
always  less  than  the  other,  and  if  the  former  eon- 
tinually  increases,  and  the  latter  continually  de- 
creases, so  that  their  difference  approaches  zero  as  a 
limit,  then  the  two  variables  have  a  common  limit 
which  lies  between  them. 

Let  any  successive  values  of  the  first  variable  be  repre- 
sented by  the  measure-numbers  of  the  lines  OA,  OB,  OC  ... ; 
and  let  the  correspond- 
ing values  of  the  second     Q  ABC 
variable  be  represented 
by    the    measure-num- 
bers of  the  lines  0A%  OB',  00'  ...,  all   measured  from  the 
same  point  0. 

Since  the  first  variable  continually  increases  and  by 
hypothesis  remains  less  than  OC',  hence  the  first  variable 
has  some  limit  (54).  Since  the  second  variable  continually 
decreases  and  remains  greater  than  OC,  hence  the  second 
variable  has  some  limit  (55).  These  two  limits  are  equal ; 
for  if  not,  the  difference  of  the  two  variables  could  not  be 
made  less  than  the  difference  of  the  two  limits,  contrary  to 
the  hypothesis.  Hence  the  two  variables  have  a  common 
limit  which  lies  between  them. 

57.  Principle  5.  If  there  are  any  two  variables, 
one  of  which  is  never  greater,  and  the  other  never 
less,  than  a  certain  fixed  number  L,  and  if  the  dif- 
ference of  the  two  variables  tends  to  zero  as  a  limit, 
then  the  two  variables  approach  L  as  a  common  limit. 

From  the  hypothesis,  the  difference  of  either  variable  from 
L  is  not  greater  than  the  difference  of  the  two  variables,  and 
will  therefore  become  and  remain  less  than  any  assigned 
number.  Hence  by  definition  each  variable  tends  to  L  as 
a  limit. 

58.  Cor.  If  two  variables  have  a  common  lim,it,  then  any 
third  variable  that  always  lies  between  them  has  the  same  limit. 


350 


PLANE  GEOMETRY  —  BOOK    VI 


\ 

^^^^     X  ^ 

>^ 

B       C 


59.  Principle  6.  If  while  approaching  their  limits 
the  ratio  of  two  related  variables  remains  constant, 
the  ratio  of  their  limits  equals  the  same  constant. 

Take  two  parallel  „ 
lines,  OL,  O^L'  aud  let 
any  successive  values 
of  the  first  variable 
be  represented  by  the 
measure-numbers  of 
the  lines  OAj  OB,  OCj 
•••;  and  let  the  cor- 
responding values  of 
the  second  variable 
be  represented  by  the  measure-numbers  of  the  lines  O'^', 
0'j5',  O'C',  •••. 

T>     1  .,       •  OA         OB        OC 

By  hypothesis       ---  =  — —  =  — ^  =  •  • .. 
^      ^^  O'A'       O'B'       O'C' 

Hence  the  two  lines,  OL,  0'l\  are  similarly  divided  at  the 
points  A,  Bf  C,  •••,  and  A',  B',  c',-". 

Therefore  the  lines  00',  AA',  BB\  •••  meet  in  a  point  P 
(V.  72,  ex.  2).  Thus  if  a  line  starts  in  the  position  PO,  and 
turns  about  P,  it  will  in  its  successive  positions  mark  off  on 
OL  and  O'Z'  corresponding  values  of  the  two  variables. 

Let  the  first  variable  have  the  limit  OL,  and  let  PL  meet 
O'Z'  in  the  point  L\ 

Then  O'V  is  the  limit  of  the  second  variable.  For  since 
OL  is  the  limit  of  the  first  variable,  hence  the  revolving  line 
can  come  as  close  to  PL  as  desired,  therefore  its  intersection 
with  o'l'  can  come  as  close  to  L'  as  desired;  thus  O'z'  is  the 
limit  of  the  second  variable. 

Therefore  the  ratio  of  the  limits  is  the  same  as  the  ratio 
of  the  variables. 

Length  of  a  Circle 

60.  The  length  of  a  straight  line  has  been  defined  as  its 
measure-number  in  terms  of  a  certain  standard  straight  line. 


MEASUREMENT  OF  THE  CIRCLE  351 

Such  measurement  presupposes  the  possibility  of  the  super- 
position of  the  standard  line,  or  of  some  of  its  submultiples, 
on  the  line  to  be  measured.  Hence  the  word  "  length  "  has  as 
yet  no  meaning  when  applied  to  a  curved  line.  The  phrase 
"  length  of  a  circle  "  will  now  be  given  a  precise  definition. 

61.  Definition.  If  in  a  circle  a  series  of  convex  polygons 
of  3,  4,  5,  ...,  sides  are  inscribed,  the  limit  approached 
by  the  length  of  the  successive  perimeters,  as  the  number 
of  sides  is  continually  increased  and  each  side  tends  to  zero 
as  a  limit,  is  called  the  length  of  the  circle. 

62.  To  justify  this  definition  it  is  necessary  to  prove  that 
this  series  of  perimeters  has  a  limit,  and  that  this  limit  is 
the  same  by  whatever  law  the  sides  tend  to  zero. 

It  will  first  be  proved  that  there  is  a  limit  when  the  suc- 
cessive inscribed  polygons  are  regular  and  when  the  number 
of  sides  is  continually  doubled.  It  will  then  be  proved  that 
the  same  limit  is  obtained  whatever  the  law  of  inscription 
may  be. 

63.  Theorem  6.  The  lengths  of  the  perimeters  of 
two  similar  regular  polygons,  one  inscribed,  the  other 
circumscribed,  to  a  given  circle,  tend  to  a  com- 
mon limit,  when  the  number  of  sides  is  continually 
doubled. 

The  ratio  of  the  perimeters  of  these  two  polygons  equals 
the  ratio  of  their  apothems  (V.  132),  and  hence  equals  the 
ratio  of  the  apothem  of  the  inscribed  polygon  to  the  radius. 

But  the  latter  ratio  tends  to  unity  as  a  limit  (53),  hence 
the  former  ratio  tends  to  unity  as  a  limit  (51).  Therefore 
the  difference  of  the  perimeters  tends  to  zero  as  a  limit  (52). 

Now  as  the  number  of  sides  is  continually  doubled  the 
circumscribed  perimeter  continually  diminishes  and  the 
inscribed  perimeter  continually  increases  (V.  135). 

Therefore  the  two  variable  perimeters  have  a  common 
limit  which  lies  between  them  (56). 


352  PLANE  GEOMETRY — BOOK   VI 

64.  Theorem  7.  The  length  of  blie  perimeter  of  any 
convex  polygon,  inscribed  or  circumscribed,  tends  to 
one  and  the  same  limit,  by  whatever  law  eojch  side 
tends  to  zero  as  a  limit. 

Let  ahc  be  a  convex  inscribed  polygon,  ABG  the  convex 
circumscribed  polygon  formed  by  the  tangents  drawn  at  the 
points  Qf  b,  c,  •••.  Let  ^)  and  P  be  the  perimeters  of  these 
polygons.  Let  L  be  the  limit  obtained  when  the  law  is  that  of 
the  preceding  theorem. 

To  prove  that^;  and  P  A^ 

have  the  common  limit 
L  by  whatever  law  the 
sides  of  each  polygon 
tend  to  zero  as  a  limit. 

The  inscribed  perim- 
eter p  is  less  than  any 
of  the  circumscribed  perimeters  considered  in  the  preceding 
theorem,  hencep  never  exceeds  the  limit  L  of  those  perimeters. 

Again,  the  circumscribed  perimeter  P  is  greater  than  any 
of  the  inscribed  perimeters  considered  in  the  preceding 
theorem  ;  hence  P  never  becomes  less  than  the  limit  L  of 
those  inscribed  perimeters. 

It  will  next  be  proved  that  p  can  come  as  near  to  P  as 
desired.     Let  the  length  of  the  radius  be  denoted  by  R. 

Since  OA  bisects  ah  at  right  angles,  hence 
aA-\-  Ah  _aA  _  R 
ah      ~  aM~  OM^ 
also  hB^Bc  ^  ^    ^^^  g^  ^^  py  ^g 

he  ON  ■- 

By  combining  the  numerators  and  denominators  of  the  frac- 
tions on  the  left,  another  fraction  is  formed  which  (by  a  theo- 
rem in  algebra)  lies  between  the  greatest  and  least  of  these 

P 
fractions.     Therefore  the  fraction  —  lies  between  the  greatest 

R         R       P 

and  least  of  the  fractions  — ,  — ,  •  •  • 
OM    ON 


MEASUREMENT  OF  THE  CIRCLE  353 

Now,   when   each   side   of  the   polygons    is   continually 

diminished,  the  fractions  — ,  — ,  •••  each  tend  to  unity 

T     -4.  /Kox  OM    ON  ^ 

as  a  limit  (53). 

Hence  the  fraction  — ,  which  lies  between  two  of  them, 

tends  to  unity  as  a  limit  (58) ;  and  therefore  the  difference 
P  —p  tends  to  zero  as  a  limit  (52). 

But  P  never  becomes  less  than  L,  and  p  never  becomes 
greater  than  L ;  hence  L  is  the  common  limit  of  P  andp  (57). 

65-  Note.  It  should  be  observed  that,  mider  the  uiost  general  law 
of  approach  now  supposed,  it  is  not  necessary  that  P  should  continually 
decrease,  nor  thatp  should  continually  increase.  Hence  57  has  been 
used  instead  of  56,  which  was  properly  employed  in  the  last  proposition. 

66.  Theorem  8.  The  lengths  of  any  two  circles  have 
the  same  ratio  as  the  radii. 

Outline.  In  the  two  circles  inscribe  similar  regular  poly- 
gons ;  and  apply  V.  132.  Imagine  the  number  of  sides  to 
be  increased ;  and  apply  59,  61. 

67.  Cor.  I.  TJie  ratio  of  the  length  of  the  circle  to  that 
of  the  diameter  is  the  same  for  all  circles. 

Apply  alternation  to  QtQ. 

Note.  The  number-correspondent  of  this  constant  ratio 
is  denoted  by  the  Greek  letter  tt.  Thus,  tt  is  the  quotient 
of  the  length  of  the  circle  by  the  length  of  the  diameter. 
The  length  of  a  circle  is  called  the  circu7nference. 

68.  Cor.  2.  If  R  is  the  length  of  the  radius,  and  C  the 
circumference,  then  C  =2-^  -  R. 

COMPUTATION    OF    THE    NUMBER   TT 

69.  The  successive  decimal  proximates  to  the  ratio  of  the 
circumference  to  the  diameter  may  be  computed  as  follows : 

Take  the  perimeters  of  some  regular  inscribed  and  the 
corresponding  circumscribed  polygons  as  found  in  45.  Com- 
pute the  perimeters  of  regular  inscribed  and  circumscribed 
polygons  of  double  the  number  of  sides  by  46.     From  these 


354  PLANE  GEOMETRY  —  BOOK   VI 

in  turn  compute  the  perimeters  of  polygons  of  double  the 
number  of  sides;  and  so  on.  These  successive  perimeters 
will  be  closer  and  closer  approximations  to  the  length  of 
the  circle.  E.g.,  if  a  decimal  proximate  of  the  fourth  order 
is  required,  continue  the  process  until  the  expressions  for 
the  inscribed  and  circumscribed  perimeters  agree  to  the 
third  decimal  place. 

For  convenience  take  the  diameter  as  standard  line; 
then  its  measure-number  is  unity. 

The  perimeters  of  the  inscribed  and  circumscribed  squares 
and  octagons  have  been  found  to  be 

p,  =  4V2  .  r  =  4:V2  .  J  ==  2.8284271, 
P4  =  8r  =  4, 


Ps  =  8  V2-V2  .  r  =  3.0614675, 
P,  =  16  (V2  -  1)  .  r  =  3.3137085. 
Now,  to  compute  P^,  use  the  result  of  V.  137,  viz. 

Pn  +  Pn_2Pn 


Pn 

"^-' 

whicl 

1  gives 

P2n  = 

Pn-\-Pn 

hence 

1 

^16  = 

_  2/)«P«  _ 

i>8  +  ^8 

--  3.1825979. 

To 

compute 

Pie,  use  the  result  of  V.  136,  viz. 

P2n 

/>2n 

=p.: 

i.e. 

PlB- 

=  Pn'P2n, 

hence 

=  Vps  •  ^16 

=  3.1214452. 

For  polygons  of  32  sides 

^32  = 

2puP^^ 
Pie  +  Pie 

=  3.1517249, 

i>32  =  Vpie  •  ^82  =  3.1365485. 


MEASUREMENT  OF  THE  CIRCLE 


355 


The  results  obtained  by  continuing  this  process  for  eight 
more  steps  are  shown  in  the  following  table : 


Number 

Perimeter  of 

Perimeter  of 

of  sides 

inscribed  polygon 

,  circumscribed  polygon 

4 

2.8284271 

4.0000000 

8 

3.0614675 

3.3137085 

16 

3.1214452 

3.1825979 

32 

3.1365485 

3.1517249 

64 

3.1403312 

3.1441184 

128 

3.1412773 

3.1422236 

266 

3.1415138 

3.1417504 

512 

3.1415729 

3.1416321 

1024 

3.1415877 

3.1416025 

2048 

3.1415914 

3.1415951 

4096 

3.1415923 

3.1415933 

8192 

3.1415926 

3.1415928 

The  last  numbers  show  that  the  length  of  the  circle 
whose  diameter  is  unity  lies  between  3.1415926  and 
3.1415928.  Hence,  the  value  tt  =  3.1415927  has  an  error  of 
less  than  one  unit  in  the  seventh  decimal  place. 

Archimedes  (250  b.c.)  obtained  the  value  Y»  which  is  correct  to 
two  decimal  places.  Metius  of  Holland  (1600  a.b.)  gave  fff,  correct 
to  six  places.  More  recently  by  methods  of  the  Calculus,  ir  has  been 
computed  to  several  hundred  figures.  Lambert  (1750  a.d.)  proved 
that  T  is  an  irrational  number  (14).  Lindemann  (1882)  proved  it 
transcendental,  i.e.  not  expressible  by  a  finite  combination  of  radicals. 


LENGTH    OF    A    CIRCULAR   ARC 

70.  Definition.  The  length  of  a  circular  arc  is  defined  as 
the  limit  to  which  tend  the  perimeters  of  any  inscribed  (or 
circumscribed)  convex  broken  line  when  each  side  tends  to 
zero  as  a  limit. 

The  existence  of  a  unique  limit  is  proved  by  the  method 
employed  in  the  two  preceding  theorems,  i.e.  by  first  con- 
sidering the  case  of  a  regular  inscribed  broken  line,  and  the 


356  PLANE  GEOMETRY  —  BOOK   VI 

corresponding  circumscribed  line,  the  number  of  sides  being 
continually  doubled ;  and  from  this  case  passing  to  the  most 
general  law  of  approach. 

71.  Cor.  I.  The  length  of  any  arc  of  a  circle  is  greater  than 
the  length  of  its  chord.     (Use  I.  89,  and  VI.  55,  ex.) 

72.  Cor.  2.  Tlie  length  of  any  arc  of  a  circle  is  less  than 
the  length  of  any  broken  line  exterior  to  it  and  Jiaving  the 
same  extremities. 

For  a  continually  decreasing  series  of  circumscribed  broken 
lines  can  be  constructed,  all  less  than  the  given  broken  line, 
hence  their  limit  is  less  than  the  same  line. 

73.  Cor.  3.     Equal  arcs  have  equal  lengths, 

74.  Cor.  4.  In  equal  circles,  according  cw  one  arc  is  greater 
than,  equal  to,  or  less  than  another,  so  is  the  length  of  the  first 
arc  greater  than,  equal  to,  or  less  than  the  length  of  the  second. 

75.  Cor.  5.  In  equal  circles,  the  ratio  of  any  two  ai'cs  is 
equal  to  the  ratio  of  the  lengths  of  the  arcs. 

Take  any  equimultiples  of  the  antecedents,  and  any  equi- 
multiples of  the  consequents,  and  apply  74. 

Area  of  a  Circle 

76.  The  definition  of  the  measure-number  of  a  polygon  in 
terms  of  a  standard  polygon  presupposes  the  possibility  of 
the  superposition  of  their  parts  by  some  mode  of  dissection. 

As  this  is  not  possible  when  the  boundary  of  the  figure 
to  be  measured  is  a  curved  line,  it  becomes  necessary  to 
give  a  precise  definition  to  the  phrase  "  area  of  a  circle." 

77.  Definition.  The  limit  to  which  the  area  of  a  polygon 
inscribed  in  a  circle  tends,  when  each  side  tends  to  zero 
as  a  limit,  is  called  the  area  of  the  circle. 

To  justify  this  definition,  it  is  necessary  to  prove  that 
there  is  such  a  limit,  and  that  its  value  is  the  same  by  what- 
ever law  each  side  approaches  zero. 


MEASUREMENT  OF  THE  CIRCLE  357 

78.  Theorem  9.  The  areas  of  two  similar  regular 
polygons,  one  inscribed,  the  other  circumscribed,  to  a 
given  circle,  tend  to  a  common  limit  when  the  num- 
ber of  sides  is  continually  doubled. 

The  ratio  of  the  areas  equals  the  ratio  of  the  squares  of 
their  apothems,  and  hence  equals  the  ratio  of  the  square  of 
the  apothem  of  the  inscribed  polygon  to  the  square  of  the 
radius.    (V.  132, 133 ;  III.  136.)    Conclude  the  proof  as  in  64. 

79.  Theorem  10.  The  area  of  any  convex  polygon, 
inscribed  or  circumscribed,  tends  to  the  same  limit  by 
whatever  law  ea^ch  side  tends  to  zero  as  a  limit. 

Use  the  figure  of  art.  64 ;  and  let  a  and  A  be  the  areas  of 
the  two  polygons.  Let  S  be  the  limit  obtained  when  the 
law  is  that  of  the  preceding  theorem. 

Show  as  in  64  that  a  never  becomes  greater  than  S,  and 
that  A  never  becomes  less  than  S.     Prove  that 
OaAh       i?2       055  c       R^ 


Oah       OM^'      Obc       ON^' 


[V.  107 


Hence  prove  that  -  tends  to  unity  as  a  limit ;  and  finally 
a 

that  S  is  the  common  limit  of  A  and  a. 

80.  Theorem  11.  The  area  of  a  circle  equals  the 
product  of  the  circumference  by  half  the  radius. 

Draw  a  regular  circumscribed  polygon. 

Its  apothem  is  equal  to  the  radius.  Hence  the  area  A  of 
the  polygon  equals  the  product  of  its  perimeter  P  by  half 
the  radius ;  that  is,         A  =  ^R'P,  [III.  137 

A      R 

or  —  =  t:- 

P     2 

On  continually  increasing  the  number  of  sides,  A  tends  to 
the  area  of  the  circle  (s),  and  P  to  the  length  (c).     [79,  64 

Since  the  quotient  of  the  variables  A  and  P  is  constant, 
the  quotient  of  their  limits  equals  the  same  constant  (59). 

Therefore  -  =  -,  and  S=l  R-C. 

C     2' 


358  PLANE  GEOMETRY — BOOK  VI 

81.  Cor.      S=2  7rR'iR  =  TrI^. 

82.  Definition.  The  area  of  a  sector  of  a  circle  is  defined 
as  the  limit  to  which  the  area  of  an  inscribed  polygonal 
sector  tends,  when  each  side  of  the  corresponding  inscribed 
broken  line  tends  to  zero. 

The  existence  of  this  limit  is  established  as  in  78,  79. 

The  area  of  a  segment  of  a  circle  may  be  defined  in  a 
similar  way.  It  is  equal  either  to  the  difference  or  the  sum 
of  the  areas  of  the  corresponding  sector  and  the  correspond- 
ing triangle,  according  as  the  arc  of  the  segment  is  less  or 
greater  than  a  semicircle. 

83.  Theorem  12.  The  area  of  a  circular  sector 
equals  the  product  of  the  length  of  its  arc  hy  half 
the  radius.     (Prove  as  in  80.) 

Ex.  Show  how  to  find  the  area  of  any  portion  of  a  plane  bounded 
by  either  straight  lines  or  by  arcs  of  circles. 

Measurement  of  Angles 

84.  The  standard  unit  for  measuring  angles  is  the  right 
angle. 

As  it  is  too  large  for  convenient  use,  a  certain  fraction  of 
it,  called  a  degree^  is  employed  in  practice. 

A  degree  is  defined  as  ^  of  a  right  angle. 

Hence  a  right  angle  equals  90  degrees,  written  90°;  a 
straight  angle  equals  180°;  and  a  perigon  equals  360°. 

Ex.  1.  How  many  degrees  are  there  in  the  sum  of  the  angles  of  a 
triangle  ?  In  the  angle  of  an  equilateral  triangle  ?  In  each  of  the 
angles  of  an  isosceles  right  triangle  ? 

Ex.  2.  How  many  degrees  are  there  in  the  angle  of  a  regular  penta- 
gon ?    A  regular  hexagon  ?    A  regular  octagon  ? 

The  sixtieth  part  of  a  degree  is  called  a  minute,  written 
1';  and  the  sixtieth  part  of  a  minute  is  called  a  second^ 
written  1". 

E.g.  the  seventh  part  of  a  right  angle  equals  12°  61'  251^". 


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